chapter 17 reaction kinetics...reaction kinetics 561 section 1 o bjectives explain the concept of...

Reaction KineticsChemists can determine the rates

at which chemical reactions occur.

C H A P T E R 1 7

The Thermite ReactionThe Thermite Reaction

R E A C T I O N K I N E T I C S 561

SECTION 1

OBJECTIVES

Explain the concept of reaction mechanism.

Use the collision theory tointerpret chemical reactions.

Define activated complex.

Relate activation energy to enthalpy of reaction.

The Reaction Process

B y studying many types of experiments, chemists have found thatchemical reactions occur at widely differing rates. For example, in thepresence of air, iron rusts very slowly, whereas the methane in naturalgas burns rapidly. The speed of a chemical reaction depends on theenergy pathway that a reaction follows and the changes that take placeon the molecular level when substances interact. In this chapter, youwill study the factors that affect how fast chemical reactions take place.

Reaction Mechanisms

If you mix aqueous solutions of HCl and NaOH, an extremely rapidneutralization reaction occurs, as shown in Figure 1.

The reaction is practically instantaneous; the rate is limited only by thespeed with which the H3O+ and OH− ions can diffuse through the waterto meet each other. On the other hand, reactions between ions of thesame charge and between molecular substances are not instantaneous.Negative ions repel each other, as do positive ions. The electron cloudsof molecules also repel each other strongly at very short distances.Therefore, only ions or molecules with very high kinetic energy canovercome repulsive forces and get close enough to react. In this section,we will limit our discussion to reactions between molecules.

Colorless hydrogen gas consists of pairs of hydrogen atoms bondedtogether as diatomic molecules, H2. Violet-colored iodine vapor is alsodiatomic, consisting of pairs of iodine atoms bonded together as I2 mol-ecules. A chemical reaction between these two gases at elevated tem-peratures produces hydrogen iodide, HI, a colorless gas. Hydrogeniodide molecules, in turn, tend to decompose and re-form hydrogenand iodine molecules, producing the violet gas shown in Figure 2. Thefollowing chemical equations describe these two reactions.

H2(g) + I2(g) ⎯→ 2HI(g)

2HI(g) ⎯→ H2(g) + I2(g)

Such equations indicate only which molecular species disappear as aresult of the reactions and which species are produced. They do notshow the reaction mechanism, the step-by-step sequence of reactions bywhich the overall chemical change occurs.

H3O+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) ⎯→ 2H2O(l) + Na+(aq) + Cl−(aq)

FIGURE 1 As NaOH solution ispoured into HCl solution, a veryrapid neutralization reaction occurs.Excess NaOH turns the phenolph-thalein indicator pink.

Although only the net chemical change is directly observable formost chemical reactions, experiments can often be designed that sug-gest the probable sequence of steps in a reaction mechanism. Each reac-tion step is usually a simple process. The equation for each steprepresents the actual atoms, ions, or molecules that participate in thatstep. Even a reaction that appears from its balanced equation to be asimple process may actually be the result of several simple steps.

For many years, the formation of hydrogen iodide, as shown in Figure2, was considered a simple one-step process. It was thought to involve theinteraction of two molecules, H2 and I2, in the forward reaction and twoHI molecules in the reverse reaction. Experiments eventually showed,however, that a direct reaction between H2 and I2 does not take place.

Alternative mechanisms for the reaction were proposed based onthe experimental results. The steps in each reaction mechanism had toadd together to give the overall equation. Note that two of the speciesin the mechanism steps—I and H2I—do not appear in the net equation.Species that appear in some steps but not in the net equation are knownas intermediates. (Notice that they cancel each other out in the follow-ing mechanisms.) The first possible mechanism has the following two-step pathway.

Step 1: I2⎯→←⎯ 2I

Step 2: 2I + H2⎯→←⎯ 2HI

I2 + H2⎯→←⎯ 2HI

The second possible mechanism has a three-step pathway.

Step 1: I2⎯→←⎯ 2I

Step 2: I + H2⎯→←⎯ H2I

Step 3: H2I + I ⎯→←⎯ 2HII2 + H2

⎯→←⎯ 2HI

The reaction between hydrogen gas and iodine vapor to produce hydro-gen iodide gas is an example of a homogeneous reaction, a reactionwhose reactants and products exist in a single phase—in this case, the gasphase.This reaction system is also an example of a homogeneous chemi-cal system because all reactants and products in all intermediate stepsare in the same phase.

Collision Theory

In order for reactions to occur between substances, their particles (mol-ecules, atoms, or ions) must collide. Furthermore, these collisions mustresult in interactions. The set of assumptions regarding collisions andreactions is known as collision theory. Chemists use this theory to inter-pret many of their observations about chemical reactions.

C H A P T E R 1 7562

FIGURE 2 Colorless hydrogeniodide gas, HI, decomposes into col-orless hydrogen gas and violetiodine gas. Some of the hydrogen gasand iodine gas will reform HI.

Consider what might happen on a molecular scale in one step of ahomogeneous reaction system. We will analyze a proposed first step ina hypothetical decomposition reaction.

AB + AB ⎯→←⎯ A2 + 2B

According to the collision theory, the two AB molecules must collide inorder to react. Furthermore, they must collide with a favorable orienta-tion and with enough energy to merge the valence electrons and disruptthe bonds of the AB molecules. If they do so, a reshuffling of bondsleads to the formation of the products, one A2 molecule and two Batoms. An effective collision is modeled in Figure 3a.

If a collision is too gentle, the two molecules simply rebound from eachother unchanged. This effect is illustrated in Figure 3b. Similarly, a col-lision in which the reactant molecules have an unfavorable orientationhas little effect. The colliding molecules rebound without reacting. Acollision that has poor orientation is shown in Figure 3c.

A chemical reaction produces new bonds which are formed betweenspecific atoms in the colliding molecules. Unless the collision brings thecorrect atoms close together and in the proper orientation, the moleculeswill not react. For example, if a chlorine molecule collides with the oxygenend of a nitrogen monoxide molecule, the following reaction may occur.

NO(g) + Cl2(g) ⎯→ NOCl(g) + Cl(g)

This reaction will not occur if the chlorine molecule strikes the nitro-gen end of the molecule.

Thus, collision theory provides two reasons why a collision betweenreactant molecules may fail to produce a new chemical species: the col-lision is not energetic enough to supply the required energy, or the col-liding molecules are not oriented in a way that enables them to reactwith each other.


AB AB

AB AB

Collisiontoo gentle

(b)

AB AB

ABAB

Collision that haspoor orientation

(c)

AB

A2

B B

AB

Effective collision,favorable orientation and energy

(a)

FIGURE 3 Three possible colli-sion patterns for AB molecules areshown. Not every collision producesa chemical reaction.

Activation Energy

Consider the reaction for the formation of water from the diatomicgases oxygen and hydrogen according to the following equation.

2H2(g) + O2(g) ⎯→ 2H2O(l)

The enthalpy of formation is quite high: ΔH 0f = −285.8 kJ/mol at

298.15 K. The free-energy change is also large: ΔG0 = −237.1 kJ/mol.Why, then, don’t oxygen and hydrogen combine spontaneously andimmediately to form water when they are mixed at room temperature?

Hydrogen and oxygen gases exist as diatomic molecules. When themolecules approach each other, the electron clouds repel each other, sothe molecules might not meet. For a reaction to occur, the colliding mol-ecules must have enough kinetic energy to intermingle the valence elec-trons. In other words, the bonds of these molecular species must bebroken in order for new bonds to be formed between oxygen and hydro-gen atoms. Bond breaking is an endothermic process, and bond formingis exothermic. Even though the net process for forming water is exother-mic, an initial input of energy is needed to overcome the repulsion forcesthat occur between reactant molecules when they are brought very closetogether. This initial energy input activates the reaction.

Once an exothermic reaction is started, the energy released isenough to sustain the reaction by activating other molecules. Thus, thereaction rate keeps increasing. It is limited only by the time required for

reactant particles to acquire the energy and makecontact. Energy from an outside source may startexothermic reactants along the pathway of reac-tion. A generalized reaction pathway for anexothermic reaction is shown as the forward reac-tion in Figure 4. The minimum amount of energyneeded to activate this reaction is the activationenergy represented by Ea . Activation energy is theminimum energy required to transform the reactantsinto an activated complex.

The reverse reaction, decomposition of watermolecules, is endothermic because the water mole-cules lie at an energy level lower than that of thehydrogen and oxygen molecules. The water mole-cules require a larger activation energy before theycan decompose to re-form oxygen and hydrogen.The energy needed to activate an endothermicreaction is greater than that required for the origi-nal exothermic change and is represented by Ea′ inFigure 4. The difference between Ea′ and Ea isequal to the energy change in the reaction, ΔE. Thisenergy change has the same numerical value for theforward reaction as it has for the reverse reactionbut with the opposite sign.

C H A P T E R 1 7564

Course of reaction

Ener

gy

Reactants

Products

Forwardreaction

(exothermic)

Reversereaction

(endothermic)

ΔE

Ea′

Activated complex

Reaction Pathways for Forward and Reverse Reactions

Ea

FIGURE 4 The difference between the activation energiesfor the reverse and forward reactions of a reversible reactionequals the energy change in the reaction, ΔE. The quantity forΔE is the same for both directions, but is negative for theexothermic direction and positive for the endothermicdirection.

The Activated Complex

When molecules collide, some of their high kinetic energy is convertedinto internal potential energy within the colliding molecules. If enoughenergy is converted, molecules with suitable orientation become acti-vated. New bonds can then form. In this brief interval of bond breakageand bond formation, the collision complex is in a transition state. Somepartial bonding exists in this transitional structure. A transitional struc-ture that results from an effective collision and that persists while oldbonds are breaking and new bonds are forming is called an activatedcomplex.

Figure 5 graphically breaks down the reaction pathway of the for-mation of hydrogen iodide gas into three steps. Beginning with the reac-tants, H2 and I2, a certain amount of activation energy, Ea1, is needed toform the activated complex that leads to the formation of the interme-diates H2 and 2I. Then more activation energy, Ea2, is needed to formthe activated complex leading to the intermediates H2I and I. In orderto arrive at the final product, 2HI, another increase in activation energyis necessary, as seen by the highest peak labeled Ea3.

An activated complex is formed when an effective collision raises theinternal energies of the reactants to their minimum level for reaction, asin Figure 4. Both forward and reverse reactions go through the same acti-vated complex. A bond that is broken in the activated complex for theforward reaction must be re-formed in the activated complex for thereverse reaction. Observe that an activated complex occurs at a high-energy position along the reaction pathway.

The kinetic-molecular theory states that the speeds and therefore thekinetic energies of the molecules increase as thetemperature increases. An increase in speed causesmore collisions, which can cause an increase in thenumber of reactions. However, an increase in thereaction rate depends on more than simply the num-ber of collisions, as Figure 3 illustrates.The collisionsbetween molecules must possess sufficient energy toform an activated complex or a reaction will nottake place. Raising the temperature of a reactionprovides more molecules that have this activationenergy and causes an increase in the reaction rate.

In its brief existence, the activated complexhas partial bonding that is characteristic of bothreactant and product. It may then re-form theoriginal bonds and separate back into the reac-tant particles, or it may form new bonds and sep-arate into product particles. The activatedcomplex, unlike the relatively stable intermediateproducts, is a very short-lived molecular complexin which bonds are in the process of being brokenand formed.


Course of reaction

Ener

gy

Ea1

Ea2

Ea3

**

*

H2 + I22HI

H2I+ I

H2+ 2I

ΔE

Activation Energy Peaks in the Formation of Activated Complexes

FIGURE 5 This energy profile graphically shows the forma-tion of three activated complexes (*) during the gas-phasereaction H2 + I2 ⎯→ 2HI.

C H A P T E R 1 7566

The energy level of reactants is always at the left-hand end of such a curve, and the energylevel of products is always at the right-hand end. The energy change in the reaction, ΔE, is thedifference between these two energy levels. The activation energy differs in the forward andreverse directions. It is the minimum energy needed to achieve effective reaction in eitherdirection. As Ea, it is the difference between the reactant energy level and the peak in thecurve. As Ea′, it is the difference between the product energy level and the peak in the curve.

ΔEforward = energy of products − energy of reactantsΔEforward = 50 kJ/mol − 0 kJ/mol = +50 kJ/mol

ΔEreverse = energy of reactants − energy of productsΔEreverse = 0 kJ/mol − 50 kJ/mol = − 50 kJ/mol

Ea = energy of activated complex − energy of reactantsEa = 80 kJ/mol − 0 kJ/mol = 80 kJ/mol

Ea′ = energy of activated complex − energy of productsEa′ = 80 kJ/mol − 50 kJ/mol = 30 kJ/mol

SOLUTION

SAMPLE PROBLEM A

Copy the energy diagram below, and label the reactants, products, �E, Ea , and Ea′. Determine the valueof �Eforward , �Ereverse , Ea , and Ea′.

For more help, go to the Math Tutor at the end of this chapter.

Ener

gy

(kJ/

mo

l)

ReverseForward

80

50

0

Ener

gy

(kJ/

mo

l)

ReverseForward

80

50

0Reactants

ProductsEa

ΔE

Ea′


Go to go.hrw.com formore practice problemsthat ask you todetermine Ea and ΔE.

Keyword: HC6RXKX

1. a. Use the method shown in the sample problem to redraw andlabel the following energy diagram. Determine the value ofΔEforward, ΔEreverse, Ea, and Ea′.

b. Is the forward reaction shown in the diagram exothermic orendothermic? Explain your answer.

2. a. Draw and label an energy diagram similar to the one shown inthe sample problem for a reaction in which Ea = 125 kJ/mol andEa′ = 86 kJ/mol. Place the reactants at energy level zero.

b. Calculate the values of ΔEforward and ΔEreverse.

c. Is this reaction endothermic or exothermic? Explain youranswer.

3. a. Draw and label an energy diagram for a reaction in which Ea = 154 kJ/mol and ΔE = 136 kJ/mol.

b. Calculate the activation energy, Ea′, for the reverse reaction.

Answers in Appendix EPRACTICE

Ener

gy

(kJ/

mo

l)

ReverseForward

100

50

−50

−100

0

−150

1. What is meant by reaction mechanism?

2. What factors determine whether a molecular colli-sion produces a reaction?

3. What is activation energy?

4. What is an activated complex?

5. How is activation energy related to the energy of reaction?

6. What is the difference between an activated com-plex and an intermediate?

7. Explain why, even though a collision may haveenergy in excess of the activation energy, reactionmay not occur.

Critical Thinking

8. ANALYZING INFORMATION Which corresponds tothe faster rate: a mechanism with a small activa-tion energy or one with a large activation energy?Explain your answer.

SECTION REVIEW

C H A P T E R 1 7568

SECTION 2 Reaction Rate

T he change in concentration of reactants per unit time as a reactionproceeds is called the reaction rate. The study of reaction rates is con-cerned with the factors that affect the rate and with the mathematicalexpressions that reveal the specific dependencies of the rate on concen-tration. The area of chemistry that is concerned with reaction rates andreaction mechanisms is called chemical kinetics.

Rate-Influencing Factors

For reactions other than simple decompositions to occur, particles mustcome into contact in a favorable orientation and with enough energy foractivation. Thus, the rate of a reaction depends on the collision fre-quency of the reactants and on the collision efficiency. Any change inreaction conditions that affects the collision frequency, the collision effi-ciency, or the collision energy affects the reaction rate. At least fiveimportant factors influence the rate of a chemical reaction.

Nature of ReactantsSubstances vary greatly in their tendencies to react. For example, hydro-gen combines vigorously with chlorine under certain conditions. Underthe same conditions, it may react only weakly with nitrogen. Sodiumand oxygen combine much more rapidly than iron and oxygen undersimilar conditions. Bonds are broken and other bonds are formed inreactions. The rate of reaction depends on the particular reactants andbonds involved.

Surface AreaGaseous mixtures and dissolved particles can mix and collide freely;therefore, reactions involving them can occur rapidly. In heterogeneousreactions, the reaction rate depends on the area of contact of the reac-tion substances. Heterogeneous reactions involve reactants in two dif-ferent phases. These reactions can occur only when the two phases arein contact. Thus, the surface area of a solid reactant is an important fac-tor in determining rate. An increase in surface area increases the rate ofheterogeneous reactions.

Solid zinc reacts with aqueous hydrochloric acid to produce zincchloride and hydrogen gas according to the following equation.

Zn(s) + 2HCl(aq) ⎯→ ZnCl2(aq) + H2(g)

OBJECTIVES

Define chemical kinetics, andexplain the two conditionsnecessary for chemical reactions to occur.

Discuss the factors that influ-ence reaction rate.

Define catalyst, and discusstwo different types.

Relate the order of a reactionto the rate law for the reaction.

Explain and write rate lawsfor chemical reactions.

www.scilinks.orgTopic: Factors Affecting

Reaction RatesCode: HC60564

This reaction occurs at the surface of the zinc solid. A cube of zinc mea-suring 1 cm on each edge presents only 6 cm2 of contact area. The sameamount of zinc in the form of a fine powder might provide a contactarea thousands of times greater than the original area. Consequently,the reaction rate of the powdered solid is much faster.

A lump of coal burns slowly when kindled in air. The rate of burningcan be increased by breaking the lump into smaller pieces, exposingmore surface area. If the piece of coal is powdered and then ignitedwhile suspended in air, it burns explosively. This is the cause of someexplosions in coal mines.

TemperatureAn increase in temperature increases the average kinetic energy of theparticles in a substance; this can result in a greater number of effectivecollisions when the substance is allowed to react with another sub-stance. If the number of effective collisions increases, the reaction ratewill increase.

To be effective, the energy of the collisions must be equal to orgreater than the activation energy. At higher temperatures, more parti-cles possess enough energy to form the activated complex when colli-sions occur.Thus, a rise in temperature produces an increase in collisionenergy as well as in collision frequency.

Decreasing the temperature of a reaction system has the oppositeeffect.The average kinetic energy of the particles decreases, so they col-lide less frequently and with less energy, producing fewer effective col-lisions. Beginning near room temperature, the reaction rates of manycommon reactions roughly double with each 10 K (10°C) rise in tem-perature. This rule of thumb should be used with caution, however. Theactual rate increase with a given rise in temperature must be deter-mined experimentally.

ConcentrationPure oxygen has five times the concentration of oxygen molecules thatair has at the same pressure; consequently, a substance that oxidizes inair oxidizes more vigorously in pure oxygen. For example, in Figure 6,the light produced when the lump of charcoal is burned in pure oxygenis much more intense than the light produced when the charcoal lumpis heated in air until combustion begins. The oxidation of charcoal is aheterogeneous reaction system in which one reactant is a gas. The reac-tion rate depends not only on the amount of exposed charcoal surfacebut also on the concentration of the reacting species, O2.

In homogeneous reaction systems, reaction rates depend on the con-centration of the reactants. Predicting the mathematical relationshipbetween rate and concentration is difficult because most chemical reac-tions occur in a series of steps, and only one of these steps determinesthe reaction rate. If the number of effective collisions increases, the rate


(a)

(b)

FIGURE 6 Carbon burns faster inpure oxygen (a) than in air (b)because the concentration of thereacting species, O2, is greater.

increases as well. In general, an increase in rate is expected if the con-centration of one or more of the reactants is increased, as depicted bythe model in Figure 7. In the system with only two molecules, shown inFigure 7a, only one collision can possibly occur. When there are fourmolecules in the system, as in Figure 7b, there can be four possible col-lisions. Under constant conditions, as the number of molecules in thesystem increases, so does the total number of possible collisionsbetween them. Figure 7c and d show a five- and eight-molecule system,allowing six and sixteen possible collisions, respectively. Lowering theconcentration should have the opposite effect. The actual effect of con-centration changes on reaction rate, however, must be determinedexperimentally.

Presence of CatalystsSome chemical reactions proceed quite slowly. Sometimes their reac-tion rates can be increased dramatically by the presence of a catalyst. Acatalyst is a substance that changes the rate of a chemical reaction with-out itself being permanently consumed. The action of a catalyst is calledcatalysis. The catalysis of the decomposition reaction of hydrogen per-oxide by manganese dioxide is shown in Figure 8. A catalyst provides analternative energy pathway or reaction mechanism in which thepotential-energy barrier between reactants and products is lowered.The catalyst may be effective in forming an alternative activated com-plex that requires a lower activation energy—as suggested in the ener-gy profiles of the decomposition of hydrogen peroxide, H2O2, shown inFigure 9—according to the following equation.

2H2O2(l) ⎯→ O2(g) + 2H2O(l)

Catalysts do not appear among the final products of reactions theyaccelerate. They may participate in one step along a reaction pathwayand be regenerated in a later step. In large-scale and cost-sensitive reac-tion systems, catalysts are recovered and reused. A catalyst that is in thesame phase as all the reactants and products in a reaction system is calleda homogeneous catalyst. When its phase is different from that of the reac-tants, it is called a heterogeneous catalyst. Metals are often used as heterogeneous catalysts. The catalysis of many reactions is promoted byadsorption of reactants on the metal surfaces, which has the effect ofincreasing the concentration of the reactants.

C H A P T E R 1 7570

(a) (b) (c) (d)

FIGURE 7 The number of mole-cules of reacting species affects thenumber of possible collisions andtherefore the reaction rate.

FIGURE 8 The reaction rate ofthe decomposition of hydrogen per-oxide, H2O2, can be increased byusing a catalyst. The catalyst usedhere is manganese dioxide, MnO2, ablack solid. A 30% H2O2 solution isadded dropwise onto the MnO2 inthe beaker and rapidly decomposesto O2 and H2O. Both the oxygen andwater appear as gases because theenergy released by the reaction caus-es much of the water to vaporize.

Rate Laws for Reactions

The relationship between the rate of a reaction and the concentrationof one reactant is determined experimentally by first keeping the con-centrations of other reactants and the temperature of the system con-stant. Then the reaction rate is measured for various concentrations ofthe reactant in question. A series of such experiments reveals how theconcentration of each reactant affects the reaction rate.

Hydrogen gas reacts with nitrogen monoxide gas at constant volumeand at an elevated constant temperature, according to the followingequation.

2H2(g) + 2NO(g) ⎯→ N2(g) + 2H2O(g)

Four moles of reactant gases produce three moles of product gases;thus, the pressure of the system diminishes as the reaction proceeds.Therate of the reaction can, therefore, be determined by measuring thechange of pressure in the vessel with time.

Suppose a series of experiments is conducted using the same initialconcentration of nitrogen monoxide but different initial concentrationsof hydrogen. The initial reaction rate is found to vary directly with thehydrogen concentration: doubling the concentration of H2 doubles therate, and tripling the concentration of H2 triples the rate. If R representsthe reaction rate and [H2] is the concentration of hydrogen in moles per


Course of reaction

Ener

gy

(kJ/

mo

l)

Heterogeneousprocess (MnO2)

Comparison of Pathways for the Decomposition of H2O2 by Various Catalysts

Ea = 75 Uncatalyzed

Ea = 58 Chemicallycatalyzed (I-)

Ea = 4 Enzyme catalyzed

Homogeneous processes

FIGURE 9 The activation energyfor a chemical reaction can bereduced by adding an appropriatecatalyst.

Chemistry in ActionGo to go.hrw.com for a full-lengtharticle on using spectroscopy tomeasure reaction kinetics.

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liter, the mathematical relationship between rate and concentration canbe expressed as follows.

R ∝ [H2]

The ∝ is a symbol that is read “is proportional to.”Now suppose the same initial concentration of hydrogen is used but

the initial concentration of nitrogen monoxide is varied.The initial reac-tion rate is found to increase fourfold when the NO concentration isdoubled and ninefold when the concentration of NO is tripled.Thus, thereaction rate varies directly with the square of the nitrogen monoxideconcentration, as described by the following proportion.

R ∝ [NO]2

Because R is proportional to [H2] and to [NO]2, it is proportional totheir product.

R ∝ [H2][NO]2

By introduction of an appropriate proportionality constant, k, theexpression becomes an equality.

R = k[H2][NO]2

An equation that relates reaction rate and concentrations of reactants iscalled the rate law for the reaction. It is applicable for a specific reactionat a given temperature. A rise in temperature increases the reactionrates of most reactions. The value of k usually increases as the tempera-ture increases, but the relationship between reaction rate and concen-tration almost always remains unchanged.

Using the Rate LawThe general form for the rate law is given by the following equation:

R = k[A]n[B]m

The reaction rate is represented by R, k is the specific rate constant,and [A] and [B] represent the molar concentrations of reactants. Therespective powers to which the concentrations are raised are representedby n and m. The rate law is applicable for a specific reaction at a given setof conditions and must be determined from experimental data.

The power to which a reactant concentration is raised is called theorder in that reactant. The value of n is said to be the order of the reac-tion with respect to [A], so the reaction is said to be “nth order in A.”Similarly, for the value of m, the reaction is said to be “mth order in B.”The orders, or powers, n and m, are usually small integers or zero. Anorder of one for a reactant means that the reaction rate is directly pro-portional to the concentration of that reactant. An order of two means

C H A P T E R 1 7572

ExplosivesIn a tiny fraction of a second, the reac-tions of explosives such as nitroglyc-erin, trinitrotoluene (TNT), anddynamite are over. These materials areprimarily organic substances containingmostly carbon, hydrogen, oxygen, andnitrogen atoms held together by rela-tively weak bonds. When “set off,”explosive materials experience rapiddecomposition. The released elementsimmediately react to form gaseous N2,CO, CO2, and NO2. The bonds in thesesmall molecules are much strongerthan those in the original explosivematerial, and so an enormous amountof energy is released. In addition, thesudden formation of gaseous materialcauses a tremendous increase in pres-sure that provides the force to demol-ish an unwanted building or breakrocks for building roads.

that the reaction rate is directly proportional to the square of the reac-tant. An order of zero means that the rate does not depend on the con-centration of the reactant, as long as some of the reactant is present. Thesum of all of the reactant orders is called the order of the reaction, oroverall order. The overall order of the reaction is equal to the sum of thereactant orders, or n + m. Some examples of observed rate laws thathave been derived experimentally are shown below. Some of these reac-tions involve nitrogen oxides, which are highly reactive gases that con-tribute to the formation of smog that can blanket an entire city, asshown in Figure 10.

3NO(g) ⎯→ N2O(g) + NO2(g) R = k[NO]2

second order in NO,second order overall

NO2(g) + CO(g) ⎯→ NO(g) + CO2(g) R = k[NO2]2

second order in NO2,zero order in CO,second order overall

2NO2(g) ⎯→ 2NO(g) + O2(g) R = k[NO2]2

second order in NO2,second order overall

2H2O2(aq) ⎯→ 2H2O(l) + O2(g) R = k[H2O2]first order in H2O2,first order overall

It is important to understand that the orders in the rate law may or maynot match the coefficients in the balanced equation. These orders mustbe determined from experimental data.

Specific Rate ConstantThe specific rate constant (k) is the proportionality constant relating therate of the reaction to the concentrations of reactants. It is important toremember the following about the value of k:

1. Once the reaction orders (powers) are known, the value of k must bedetermined from experimental data.

2. The value of k is for a specific reaction; k has a different value forother reactions, even at the same conditions.

3. The units of k depend on the overall order of the reaction.4. The value of k does not change for different concentrations of reac-

tants or products. So, the value of k for a reaction remains the samethroughout the reaction and does not change with time.

5. The value of k is for the reaction at a specific temperature; if weincrease the temperature of the reaction, the value of k increases.

6. The value of k changes (becomes larger) if a catalyst is present.


FIGURE 10 A cloud of pollutedair, commonly known as smog, set-tles over a city. Smog is common inindustrialized areas, where highlyreactive gases and particulate matterare released into the air.

C H A P T E R 1 7574

The general rate law for this reaction has the form R = k[HI]n. We need to deduce the valueof the power n.

Find the ratio of the reactant concentrations between two experiments, such as 1 and 2, ⎯[[HH

II]]2

1⎯.

Then, see how the ratio of concentration affects the ratio of rates, ⎯RR

2

1⎯.

Concentration ratio: = = 2.0; rate ratio: = = 4.0

Thus, when the concentration changes by a factor of 2, the rate changes by a factor of 4, or22, so the rate law is R = k[HI]2.

To find the value of k, we can rearrange the rate law and substitute known values for anyone experiment. Do the following for Experiment 1:

k = = = 4.9 M−1s−1

By comparing items 1 and 3 in the table, we see that when [HI] is tripled, the rate changesby a factor of 9, or 32. This rate change confirms that the order is 2. The same value of k canbe calculated from any other experiment. Thus, the rate law and k are correct.

1.1 × 10−3 M/s⎯⎯

(0.015 M)2

R⎯[HI]2

4.4 × 10−3 M/s⎯⎯1.1 × 10−3 M/s

R2⎯R1

0.030 M⎯0.015 M

[HI]2⎯[HI]1

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

4 EVALUATE

SAMPLE PROBLEM B

Three experiments that have identical conditions were performed to measure the initial rate of the reaction

2HI(g) ⎯→ H2(g) + I2(g)

The results for the three experiments in which only the HI concentration was varied are as follows:

Write the rate law for the reaction. Find the value and units of the specific rate constant.


Experiment [HI] (M) Rate (M/s)

1 0.015 1.1 × 10−3

2 0.030 4.4 × 10−3

3 0.045 9.9 × 10−3

Go to go.hrw.com formore practice problemsthat ask you todetermine rate law andrate constant.

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1. For the reaction 3A ⎯→ C, the initial concentration of A was 0.2M and the reaction rate was 1.0 M/s. When [A] was doubled, thereaction rate increased to 4.0 M/s. Determine the rate law for thereaction.

2. The rate law for a reaction is found to be rate = k[X]3. By whatfactor does the rate increase if [X] is tripled?



The general rate law for this reaction has the form R = k[A]n[B]m. We need to calculate thevalues of the powers n and m.

Find the ratio of the reactant concentrations between two experiments that have the same[A] but different [B]. Then, see how this ratio affects the ratio of rates, ⎯R

R2

1⎯; this ratio of rates

lets us find the value of m. A similar approach of comparing two experiments that have thesame [B] but a different [A] lets us find the value of n.

First compare Experiments 1 and 2, which have the same [A], to find m:


Thus, when the concentration of B changes by a factor of 2, the rate changes by a factor of2, or 21. So, m is 1, and the reaction is first order in B.

Then, compare Experiments 1 and 3, which have the same [B], to find n:


Thus, when the concentration of A changes by a factor of 3, the rate changes by a factor of9, or 32. So, n is 2, and the reaction is second order in A.

The rate law is R � k[A]2[B].

To find the value of k, we can rearrange the rate law and substitute known values for anyone experiment. Do the following for Experiment 1:

k = = = 2.3 × 10−8 M−2s−1

The same value of k can be calculated from the data for any other experiment. So, the ratelaw and the calculation of k are correct.

8.0 × 10−8M/s⎯⎯(1.2 M)2(2.4 M)

R⎯[A]2[B]

7.2 × 10−7 M/s⎯⎯8.0 × 10−8 M/s

R3⎯R1

3.6 M⎯1.2 M

[A]3⎯[A]1

8.0 × 10−8 M/s⎯⎯4.0 × 10−8 M/s

R1⎯R2

2.4 M⎯1.2 M

[B]1⎯[B]2

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

4 EVALUATE

SAMPLE PROBLEM C

Three experiments were performed to measure the initial rate of the reaction

A + B ⎯→ C

Conditions were identical in the three experiments, except that the concentrations of reactants varied. Theresults are as follows:

Write the rate law for the reaction. Find the value and units of the specific rate constant.


Experiment [A] (M) [B] (M) Rate (M/s)

1 1.2 2.4 8.0 × 10−8

2 1.2 1.2 4.0 × 10−8

3 3.6 2.4 7.2 × 10−7

Rate Laws and Reaction PathwayThe form of the rate law depends on the reaction mechanism. For areaction that occurs in a single step, the reaction rate of that step is proportional to the product of the reactant concentrations, each ofwhich is raised to its stoichiometric coefficient. For example, supposeone molecule of gas A collides with one molecule of gas B to form twomolecules of substance C, according to the following equation.

A + B ⎯→ 2C

One particle of each reactant is involved in each collision. Thus, dou-bling the concentration of either reactant will double the collision fre-quency. It will also double the reaction rate for this step. Therefore, therate for this step is directly proportional to the concentration of A andB. The rate law for this one-step forward reaction follows.

Rforward = kforward[A][B]

Now suppose the reaction is reversible. In the reverse step, two mol-ecules of C must decompose to form one molecule of A and one of B or2C ⎯→ A + B.

Thus, the reaction rate for this reverse step is directly proportional to[C] × [C]. The rate law for the reverse step is Rreverse = kreverse[C]2.

The power to which the molar concentration of each reactant is raisedin the rate laws above corresponds to the coefficient for the reactant inthe balanced chemical equation. Such a relationship holds only if thereaction follows a simple one-step path, that is, if the reaction occurs atthe molecular level exactly as written in the chemical equation.

If a chemical reaction proceeds in a sequence of steps, the rate law isdetermined from the slowest step because it has the lowest rate. Thisslowest-rate step is called the rate-determining step for the chemicalreaction.

Consider the reaction of nitrogen dioxide and carbon monoxide.

NO2(g) + CO(g) ⎯→ NO(g) + CO2(g)

The reaction is believed to be a two-step process represented by the fol-lowing mechanism.

Step 1: NO2 + NO2 ⎯→ NO3 + NO slowStep 2: NO3 + CO ⎯→ NO2 + CO2 fast

In the first step, shown in Figure 11, two molecules of NO2collide, forming the intermediate species NO3. This moleculethen collides with one molecule of CO and reacts quickly toproduce one molecule each of NO2 and CO2. The first step isthe slower of the two and is therefore the rate-determiningstep. We can write the rate law from this rate-determiningstep, which has two molecules of NO2 as the reactants.

R = k [NO2]2

The rate law does not include [CO] because CO reacts after the rate-determining step and does not affect the rate.2NO2 NO3 NO+

FIGURE 11 This diagram is a representation of Step 1 in the reac-tion of nitrogen dioxide and carbonmonoxide. Notice the formation ofthe intermediate species NO3 aftertwo molecules of NO2 collide.

C H A P T E R 1 7576


Go to go.hrw.com formore practice problemsthat ask you todetermine rate laws andeffects on reaction rates.

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If we combine these two steps, the intermediate, F, cancels out and we are left with the original equation. The first step is the slower step, and is considered the rate-determining step. We can write the rate law from this rate-determining step.

R = k [NO2][F2]

SOLUTION

SAMPLE PROBLEM D

Nitrogen dioxide and fluorine react in the gas phase according to the following equation.

2NO2(g) + F2(g) ⎯→ 2NO2F(g)

A proposed mechanism for this reaction follows.

Step 1: NO2 + F2 ⎯→ NO2F + F (slow)Step 2: F + NO2 ⎯→ NO2F (fast)

Identify the rate-determining step, and write an acceptable rate law.


Because the equation represents a single-step mechanism, the rate law can be written from the equation (otherwise, it could not be). The rate will vary directly with the concentra-tion of X, which has an implied coefficient of 1 in the equation. And the rate will vary direct-ly with the square of the concentration of Y, which has the coefficient of 2: R = k[X][Y]2.

a. Doubling the concentration of X will double the rate (R = k[2X][Y]2).b. Doubling the concentration of Y will increase the rate fourfold (R = k[X][2Y]2).c. Using one-third the concentration of Y will reduce the rate to one-ninth of its original

value (R = k[X][ Y]2).13

SOLUTION

SAMPLE PROBLEM E

A reaction involving reactants X and Y was found to occur by a one-step mechanism:X + 2Y ⎯→ XY2. Write the rate law for this reaction, and then determine the effect of each of the following on the reaction rate:a. doubling the concentration of Xb. doubling the concentration of Yc. using one-third the concentration of Y

1. The rate of a hypothetical reaction involving L and M is found todouble when the concentration of L is doubled and to increasefourfold when the concentration of M is doubled. Write the ratelaw for this reaction.

2. At temperatures below 498 K, the following reaction takes place.

NO2(g) + CO(g) ⎯→ CO2(g) + NO(g)

Doubling the concentration of NO2 quadruples the rate of CO2formed if the CO concentration is held constant. However, dou-bling the concentration of CO has no effect on the rate of CO2formation. Write a rate-law expression for this reaction.


C H A P T E R 1 7578

Materials• Bunsen burner• paper ash• copper foil strip• graduated cylinder, 10 mL• magnesium ribbon• matches• paper clip• sandpaper• steel wool• 2 sugar cubes• white vinegar• zinc strip• 6 test tubes, 16 × 150 mm• tongs

QuestionHow do the type of reactants,surface area of reactants,concentration of reactants,and catalysts affect the ratesof chemical reactions?

ProcedureRemove all combustible materialfrom the work area. Wear safetygoggles and an apron. Record allyour results in a data table.

1. Add 10 mL of vinegar to eachof three test tubes. To one testtube, add a 3 cm piece ofmagnesium ribbon; to a sec-ond, add a 3 cm zinc strip; andto a third, add a 3 cm copperstrip. (All metals should be thesame width.) If necessary, pol-ish the metals with sandpaperuntil they are shiny.

2. Using tongs, hold a paper clipin the hottest part of the burn-er flame for 30 s. Repeat with aball of steel wool 2 cm indiameter.

3. To one test tube, add 10 mL ofvinegar; to a second, add 5 mLof vinegar plus 5 mL of water;and to a third, add 2.5 mL ofvinegar plus 7.5 mL of water.To each of the three test tubes,add a 3 cm piece of magne-sium ribbon.

4. Using tongs, hold a sugar cubeand try to ignite it with amatch. Then try to ignite it in aburner flame. Rub paper ashon a second cube, and try toignite it with a match.

Discussion

1. What are the rate-influencingfactors in each step of theprocedure?

2. What were the results fromeach step of the procedure?How do you interpret eachresult?

Factors Influencing Reaction Rate

Wear safety goggles and an apron.

1. What is studied in the branch of chemistry that isknown as chemical kinetics?

2. List five important factors that influence the rateof chemical reactions.

3. What is a catalyst? Explain the effect of a catalyston the rate of chemical reactions. How does a catalyst influence the activation energy requiredby a particular reaction?

4. What is meant by a rate law for a chemical reaction? Explain the conditions under which a rate law can be written from a chemical equation. When can a rate law not be writtenfrom a single step?

Critical Thinking

5. RELATING IDEAS Using the ideas of reaction ki-netics, explain the purpose of food refrigeration.

SECTION REVIEW


Catalytic ConvertersTo see an important example of het-erogeneous catalysis, you do notneed to look any farther than thestreets near your home. The catalyticconverter, an important part of avehicle’s exhaust system, uses metalcatalysts to reduce harmful gaseouspollutants.

In an automobile engine, hydrocar-bon molecules in gasoline or dieselfuel undergo a combustion reactionwith oxygen from air to make carbondioxide, CO2, and water. The correctstoichiometric ratio of fuel to oxygenis required for the fuel to be com-pletely burned in the reaction.Additional reaction products areformed when not enough oxygen orexcess oxygen is present. These prod-ucts include carbon monoxide, CO,and NOx compounds, such as nitricoxide, NO, and nitrogen dioxide, NO2.There is also leftover unburned fuel,which is called a volatile organiccompound (VOC).

The Clean Air Act, enacted in1990, regulates automobile emis-sions of CO, NOx, and VOCs. Withouta catalytic converter, a car wouldrelease all of the byproducts ofincomplete combustion into theatmosphere. In addition to beingharmful themselves, NOx com-pounds, CO, and VOCs react withsunlight to make ozone, O3. In thelower atmosphere, ozone is a majorpart of photochemical smog. NOxgases can also mix with rainwater toproduce acid rain.

Catalytic converters use pre-cious metal catalysts to changethe gases coming from theengine into less harmful gases.A combination of rhodium andplatinum, and sometimes palla-dium, is used to convert nitro-gen compounds back into N2 and O2.This combination also converts COinto CO2 and converts VOCs into CO2and water. The catalysts need O2from the air and temperatures aboveapproximately 500°F to work proper-ly. The temperatures are achievedfrom the normal operation of the carengine. However, until the car enginereaches the temperatures needed forthe catalysts to work, CO, NOx, andVOCs will be released into the air bythe automobile.

The interior structure of a catalyticconverter is usually made of aceramic honeycomb with a surfacecoating of metal catalyst particles.The honeycomb has many holes forthe gases to pass through and pro-vides a large surface area for themetal to be deposited on. A largesurface area is needed to maximizethe reactions that occur during het-erogeneous catalysis because thetransformation of the gas moleculesoccurs at the surface of the metal.

Up to 90% of CO, NOx, and VOCsare typically eliminated from auto-mobile exhaust by a catalytic con-verter. Although catalytic convertersare beneficial to our environment,they could still be improved.

Catalysts that work at lower temper-atures would reduce an automobile’semission during the first few minutesof operation. Other gases that areemitted by cars may also pose prob-lems for the environment. Nitrousoxide, N2O, can be formed from theincomplete reduction of NOx in cat-alytic converters. Unlike the NOxgases, N2O can travel to the upperatmosphere, where it can destroyozone. As a greenhouse gas, N2O ismore that 300 times more potentthan CO2.

1. Why do you think a heteroge-neous catalyst is used instead ofhomogeneous catalyst in a cat-alytic converter?

2. Nitrous oxide, N2O, actually hasbeneficial uses, despite its role asa greenhouse gas. Can you namea beneficial use of N2O?

Questions

The ceramic honeycomb inside a cat-alytic converter is coated with a metalcatalyst.

C H A P T E R H I G H L I G H T S

C H A P T E R 1 7580

• The rate of reaction is influenced by the following factors:nature of reactants, surface area, temperature, concentration ofreactants, and the presence of catalysts.

• The rates at which chemical reactions occur can sometimes beexperimentally measured and expressed in terms of math-ematical equations called rate laws.

• Rate laws are determined by studying how reaction ratedepends on concentration.

• The step-by-step process by which an overall chemical reactionoccurs is called the reaction mechanism.

• In order for chemical reactions to occur, the particles of thereactants must collide.

• Activation energy is needed to merge valence electrons and toloosen bonds sufficiently for molecules to react.

• An activated complex is formed when an effective collisionbetween molecules of reactants raises the internal energy tothe minimum level necessary for a reaction to occur.

reaction mechanismintermediatehomogenous reactioncollision theoryactivation energyactivated complex

reaction ratechemical kineticsheterogeneous

reactionscatalystcatalysishomogeneous

catalyst

heterogeneouscatalyst

rate laworderrate-determining step

Vocabulary

Vocabulary

The Reaction Process

Reaction Rate

C H A P T E R R E V I E W


The Reaction ProcessSECTION 1 REVIEW

1. a. What is the collision theory?b. According to this theory, what two conditions

must be met for a collision between reactantmolecules to be effective in producing newchemical species?

2. a. What condition must be met for an activatedcomplex to result from the collision of reac-tant particles?

b. Where, in terms of energy, does the activatedcomplex occur along a typical reaction pathway?

3. In a reversible reaction, how does the activationenergy required for the exothermic changecompare with the activation energy required for the endothermic change?

4. Would you expect the following equation torepresent the mechanism by which propane,C3H8, burns? Why or why not?

C3H8(g) + 5O2(g) ⎯→ 3CO2(g) + 4H2O(g)5. The decomposition of nitrogen dioxide

2NO2 ⎯→ 2NO + O2 occurs in a two-stepsequence at elevated temperatures. The firststep is NO2 ⎯→ NO + O. Predict a possible second step that, when combined with the firststep, gives the complete reaction.

Ener

gy

(kJ/

mo

l)

ReverseForward

80

60

40

20

0

−20

Ener

gy

(kJ/

mo

l)

ReverseForward

20

0

−20

−40

−60

Ener

gy

(kJ/

mo

l)

ReverseForward

70

5060

40

2030

100

a.

b.

c.

PRACTICE PROBLEMS

6. For each of the energy diagrams providedbelow, label the reactants, products, ΔE, Ea, andEa′. Also determine the values of ΔE for theforward and reverse reactions, and determinethe values of Ea and Ea′. (Hint: See SampleProblem A.)

CHAPTER REVIEW

7. Draw and label energy diagrams that depict thefollowing reactions, and determine all remainingvalues. Place the reactants at energy level zero.a. ΔEforward = −10 kJ/mol Ea′ = 40 kJ/molb. ΔEforward = −95 kJ/mol Ea = 20 kJ/molc. ΔEreverse = −40 kJ/mol Ea′ = 30 kJ/mol

Reaction RateSECTION 2 REVIEW

8. Define the rate-determining step for a chemicalreaction.

9. Write the general equation for the rate law, andlabel the various factors.

PRACTICE PROBLEMS

10. a. Determine the overall balanced equation fora reaction that has the following proposedmechanism, and write an acceptable rate law.(Hint: See Sample Problem C.)Step 1: B2 + B2 ⎯→ E3 + D slowStep 2: E3 + A ⎯→ B2 + C2 fast

b. Give the order of the reaction with respect toeach reactant.

c. What is the overall order of the reaction?11. A reaction that involves reactants A and B is

found to occur in the one-step mechanism:2A + B ⎯→ A2B. Write the rate law for thisreaction, and predict the effect of doubling theconcentration of either reactant on the overallreaction rate. (Hint: See Sample Problem C.)

12. A chemical reaction is expressed by the bal-anced chemical equation A + 2B ⎯→ C. Threereaction-rate experiments yield the followingdata.

a. Determine the rate law for the reaction.b. Calculate the value of the specific rate

constant.

c. If the initial concentrations of both A and Bare 0.30 M, at what initial rate is C formed?

d. What is the order of the reaction withrespect to A?

e. What is the order of the reaction withrespect to B?

13. Draw and label energy diagrams that depict thefollowing reactions, and determine all remainingvalues. Place the reactants at energy level zero.a. ΔE = +30 kJ/mol Ea′ = 20 kJ/molb. ΔE = −30 kJ/mol Ea = 20 kJ/mol

14. A particular reaction is found to have the fol-lowing rate law.

R = k[A][B]2

How is the rate affected by each of the follow-ing changes?a. The initial concentration of A is cut in half.b. The initial concentration of B is tripled.c. The concentration of A is doubled, but the

concentration of B is cut in half.d. A catalyst is added.

15. For each of the following pairs, choose the sub-stance or process that you would expect to reactmore rapidly.a. granulated sugar or powdered sugarb. zinc in HCl at 298.15 K or zinc in HCl at

320 Kc. 5 g of thick platinum wire or 5 g of thin plati-

num wire16. The following data relate to the reaction

A + B ⎯→ C. Find the order with respect to each reactant.

MIXED REVIEW

C H A P T E R 1 7582

Experiment Initial Initial Initial rate ofnumber [A] [B] formation of C

1 0.20 M 0.20 M 2.0 × 10−4 M/min2 0.20 M 0.40 M 8.0 × 10−4 M/min3 0.40 M 0.40 M 1.6 × 10−3 M/min

[A] (M) [B] (M) Rate (M/s)

0.08 0.06 0.0120.08 0.03 0.0060.04 0.06 0.003

CHAPTER REVIEW


17. Predicting Outcomes The balanced equationfor a rapid homogeneous reaction between twogases is as follows: 4A + B ⎯→ 2C + 2D.Because the simultaneous collision of four mol-ecules of one reactant with one molecule of theother reactant is extremely improbable, whatwould you predict about the nature of the reac-tion mechanism for this reaction system?

18. Evaluating Ideasa. How can you justify calling the reaction

pathway that is shown in Figure 4 the minimum-energy pathway for reaction?

b. What significance is associated with the maximum-energy region of this minimum-energy pathway?

19. Applying Models Explain why there is a dan-ger of explosion in places such as coal mines,sawmills, and grain elevators, where largeamounts of dry, powdered combustible materi-als are present.

20. Evaluating Methods What property wouldyou measure to determine the reaction rate forthe following reaction? Justify your choice.

2NO2(g) ⎯→ N2O4(g)

21. Look for situations around your house in whichprocesses are speeded up by an increase in temperature or slowed down by a decrease intemperature. Make a list, and discuss the differ-ent processes.

RESEARCH & WRITING

CRITICAL THINKING

22. Boilers are sometimes used to heat large build-ings. Deposits of CaCO3, MgCO3, and FeCO3can hinder the boiler operation. Aqueous solu-tions of hydrochloric acid are commonly used toremove these deposits. The general equation forthe reaction is written below.MCO3(s) + 2H3O+(aq) ⎯→

M2+(aq) + 3H2O(l) + CO2(g)In the equation, M stands for Ca, Mg, or Fe.Design an experiment to determine the effect ofvarious HCl concentrations on the rates of thisreaction. Present your design to the class.

ALTERNATIVE ASSESSMENT

Graphing CalculatorReaction Orders

Go to go.hrw.com for a graphing calculatorexercise that asks you to calculate the orderof a reaction from the reaction rates and concentrations.

Keyword: HC6RXKX

Math Tutor WRITING RATE LAWS

Factors such as surface area and temperature affect the rate of reactions because theyaffect the frequency and energy of collisions between particles. The concentrations ofreactants can also affect the frequency of collisions. If other factors are kept constant,the rates of most chemical reactions will be determined by the concentrations of reac-tants. Thus, it is possible to write an equation called a rate law that relates the rate of areaction to the concentrations of reactants.

C H A P T E R 1 7584

1. Nitrogen monoxide and oxygen react to produce nitrogen dioxide according to the following equation:

O2(g) + 2NO(g) ⎯→ 2NO2(g)Use the data in the following table to write arate law for this reaction.

2. Hydrogen reacts with ethyne, C2H2, to pro-duce ethane, C2H6, as shown below:

2H2(g) + C2H2(g) ⎯→ C2H6(g)Use the data in the following table to write arate law for this reaction.

SAMPLE

PRACTICE PROBLEMS

Fluorine gas reacts with chlorine dioxide according to the following equation.F2(g) + 2ClO2(g) ⎯→ 2FClO2(g)

Use the following experimental data to write a rate law for this reaction.

To write the rate law, first examine the data to see how the rate of reaction changes as theconcentrations of the reactants change.• When [F2] doubles and [ClO2] remains constant, the rate of reaction doubles from 1.1 ×

10−3 mol/L•s to 2.2 × 10−3 mol/L•s. So, the rate is directly proportional to [F2], or R α [F2].• When [ClO2] doubles and [F2] remains constant, the rate of reaction also doubles from

1.1 × 10−3 mol/L•s to 2.2 × 10−3 mol/L•s. So, the rate is directly proportional to [ClO2], orR α [ClO2].

• Because rate is proportional to both [F2] and [ClO2], you can write the rate law R = k[F2][ClO2]. The data from Trial 4 help confirm the rate law because when both [F2]and [ClO2] double, the rate increases by a factor of four, from 1.1 × 10−3 mol/L•s to 4.4 × 10−3 mol/L•s.

Trial Concentration of F2 Concentration of Cl02 Rate (mol/L•s)1 0.10 M 0.10 M 1.1 × 10−3

2 0.20 M 0.10 M 2.2 × 10−3

3 0.10 M 0.20 M 2.2 × 10−3

4 0.20 M 0.20 M 4.4 × 10−3

Trial [O2] [NO] Reaction Rate(mol/L•s)

1 1.20 × 10−2 M 1.40 × 10−2 M 3.30 × 10−3

2 2.40 × 10−2 M 1.40 × 10−2 M 6.60 × 10−3

3 1.20 × 10−2 M 2.80 × 10−2 M 1.32 × 10−2

Trial [H2] [C2H2] Reaction Rate(mol/L•min)

1 0.20 M 0.20 M 1.5 × 10−4

2 0.40 M 0.20 M 3.0 × 10−4

3 0.20 M 0.40 M 1.5 × 10−4


Answer the following items on a separate piece of paper.

MULTIPLE CHOICE

1.The sequence of steps that occurs in a reactionprocess is called theA. order of the reaction.B. rate law.C. overall reaction.D. reaction mechanism.

2.To be effective, a collision requiresA. enough energy only.B. favorable orientation only.C. enough energy and a favorable orientation.D. a reaction mechanism.

3.How does the energy of the activated complexcompare with the energies of the reactants andproducts?A. It is lower than both the energy of the reac-

tants and the energy of the products.B. It is lower than the energy of the reactants

but higher than the energy of the products.C. It is higher than the energy of the reactants

but lower than the energy of the products.D. It is higher than both the energy of the reac-

tants and the energy of the products.

4.If a collision between molecules is very gentle,the molecules areA. more likely to be oriented favorably.B. less likely to be oriented favorably.C. likely to react.D. likely to rebound without reacting.

5.A species that changes the rate of a reaction butis neither consumed nor changed isA. a catalyst.B. an activated complex.C. an intermediate.D. a reactant.

6.A rate law relatesA. reaction rate and temperature.B. reaction rate and concentration.C. temperature and concentration.D. energy and concentration.

7.In a graph of how energy changes with reactionprogress, the activated complex appears at theA. left end of the curve.B. right end of the curve.C. bottom of the curve.D. peak of the curve.

8.The slowest step in a mechanism is calledA. the rate-determining step.B. the uncatalyzed reaction.C. the activation step.D. None of the above

9.A certain reaction is zero order in reactant Aand second order in reactant B. What happensto the reaction rate when the concentrations ofboth reactants are doubled?A. The reaction rate remains the same.B. The reaction increases by a factor of two.C. The reaction rate increases by a factor of four.D. The reaction rate increases by a factor of eight.

SHORT ANSWER

10.Two molecules collide but bounce apartunchanged. What two reasons could account fortheir failure to react?

11.Sketch a diagram that shows how the energychanges with the progress of an endothermicreaction. Label the curve “Reactants,”“Products,” and “Activated complex.” On thesame diagram, sketch a second curve to showthe change caused by a catalyst.

EXTENDED RESPONSE

12.Suggest ways of measuring the concentration ofa reactant or product in a reaction mixture.

13.Why are reaction orders not always equal to thecoefficients in a balanced chemical equation?

Sometimes, only one part of agraph or table is needed to answer a question. Insuch cases, focus on only the information that isrequired to answer the question.

Standardized Test Prep

Rate of a Chemical Reaction

CHAPTER LAB MICRO-L A B

BACKGROUNDIn this experiment, you will determine the rate of thereaction whose net equation is written as follows:

3Na2S2O5(aq) + 2KIO3(aq) + 3H2O(l) 2KI(aq) + 6NaHSO4(aq)

One way to study the rate of this reaction is toobserve how fast Na2S2O5 is used up. After all theNa2S2O5 solution has reacted, the concentration ofiodine, I2, an intermediate in the reaction, increases.A starch indicator solution, added to the reactionmixture, will change from clear to a blue-black colorin the presence of I2.

In the procedure, the concentrations of the reac-tants are given in terms of drops of solution A anddrops of solution B. Solution A contains Na2S2O5,the starch indicator solution, and dilute sulfuric acidto supply the hydrogen ions needed to catalyze thereaction. Solution B contains KIO3. You will run thereaction with several different concentrations of thereactants and record the time it takes for the blue-black color to appear.

SAFETY

For review of safety, please see the Safety in theChemistry Laboratory in the front of your book.

PREPARATION1. Prepare a data table with six rows and six

columns. Label the boxes in the first row of thesecond through sixth columns “Well 1,” “Well 2,”“Well 3,” “Well 4,” and “Well 5.” In the first col-umn, label the boxes in the second through sixthrows “Time reaction began,” “Time reactionstopped,” “Drops of solution A,” “Drops of solu-tion B,” and “Drops of H2O.”

H+⎯→

OBJECTIVES

• Prepare and observe several different reaction mixtures.

• Demonstrate proficiency in measuringreaction rates.

• Relate experimental results to a rate lawthat can be used to predict the results ofvarious combinations of reactants.

MATERIALS

• 8-well microscale reaction strips, 2

• distilled or deionized water

• fine-tipped dropper bulbs or small microtip pipets, 3

• solution A

• solution B

• stopwatch or clock with second hand

C H A P T E R 1 7586

6. Observe the lower wells. Note the sequence inwhich the solutions react, and record the numberof seconds it takes for each solution to turn ablue-black color.

CLEANUP AND DISPOSAL7. Dispose of the solutions in the container

designated by your teacher. Wash yourhands thoroughly after cleaning up thearea and equipment.

ANALYSIS AND INTERPRETATION1. Organizing Data: Calculate the time elapsed for

the complete reaction of each combination ofsolutions A and B.

2. Evaluating Data: Make a graph of your results.Label the x-axis “Number of drops of solutionB.” Label the y-axis “Time elapsed.” Make asimilar graph for drops of solution B versus rate(1/time elapsed).

3. Analyzing Information: Which mixture reactedthe fastest? Which mixture reacted the slowest?

4. Evaluating Methods: Why was it important toadd the drops of water to the wells that containedfewer than five drops of solution B? (Hint: Figureout the total number of drops in each of thereaction wells.)

CONCLUSIONS1. Evaluating Conclusions: Which of the following

variables that can affect the rate of a reaction is tested in this experiment: temperature,catalyst, concentration, surface area, or nature of reactants? Explain your answer.

2. Applying Ideas: Use your data and graphs todetermine the relationship between the concen-tration of solution B and the rate of the reaction.Describe this relationship in terms of a rate law.

EXTENSIONS1. Predicting Outcomes: What combination of

drops of solutions A and B would you use if you wanted the reaction to last exactly 2.5 min?

FIGURE A

2. Obtain three dropper bulbs or small microtippipets, and label them “A,” “B,” and “H2O.”

3. Fill the bulb or pipet A with solution A, the bulb or pipet B with solution B, and the bulb or pipet for H2O with distilled water.

PROCEDURE1. Using the first 8-well strip, place five drops of

solution A into each of the first five wells.Record the number of drops in the appropriateplaces in your data table. For the best results,try to make all drops about the same size.

2. In the second 8-well reaction strip, place onedrop of solution B in the first well, two drops inthe second well, three drops in the third well, fourdrops in the fourth well, and five drops in thefifth well. Record the number of drops in yourdata table.

3. In the second 8-well strip that contains drops ofsolution B, add four drops of water to the firstwell, three drops to the second well, two dropsto the third well, and one drop to the fourthwell. Do not add any water to the fifth well.

4. Carefully invert the second strip. The surfacetension should keep the solutions from fallingout of the wells. Place the strip well-to-well on top of the first strip, as shown in Figure A.

5. Hold the strips tightly together as shown inFigure A, and record the exact time, or set thestopwatch, as you shake the strips once. Thisprocedure should mix the upper solutions witheach of the corresponding lower ones.


Chemical EquilibriumThe creation of stalactites and stalagmites

is the result of a reversible chemical reaction.

C H A P T E R 1 8

Carlsbad Caverns, New Mexico

C H E M I C A L E Q U I L I B R I U M 589

SECTION 1

OBJECTIVES

Define chemical equilibrium.

Explain the nature of theequilibrium constant.

Write chemical equilibriumexpressions and carry out calculations involving them.

The Nature ofChemical Equilibrium

I n systems that are in equilibrium, opposing processes occur at thesame time and at the same rate. For example, when an excess of sugar isplaced in water, some sugar molecules go into solution, and others remainundissolved. At equilibrium, molecules of sugar are crystallizing at thesame rate that molecules from the crystal are dissolving.The rate of evap-oration of a liquid in a closed vessel can eventually be equaled by the rateof condensation of its vapor. The resulting equilibrium vapor pressure isa characteristic of the liquid at the prevailing temperature.The precedingexamples are physical equilibria. In this chapter, we will expand on theconcept of equilibrium to include chemical reactions. You will learn howa system at equilibrium responds when equilibrium conditions are alteredby changing concentration, pressure, and temperature.

Reversible Reactions

Theoretically, every reaction can proceed in two directions, forward andreverse. Thus, essentially all chemical reactions are considered to bereversible under suitable conditions. A chemical reaction in which theproducts can react to re-form the reactants is called a reversible reaction.

Mercury(II) oxide decomposes when heated.

2HgO(s) 2Hg(l) + O2(g)

Mercury and oxygen combine to form mercury(II) oxide when heatedgently.

2Hg(l) + O2(g) 2HgO(s)

Figure 1 shows both of these reactions taking place. Supposemercury(II) oxide is heated in a closed container from which neither themercury nor the oxygen can escape. Once decomposition has begun, themercury and oxygen released can recombine to form mercury(II) oxideagain. Thus, both reactions can proceed at the same time. Under theseconditions, the rate of the synthesis reaction will eventually equal that ofthe decomposition reaction. At equilibrium, mercury and oxygen will

Δ⎯→

Δ⎯→

FIGURE 1 When heated, mer-cury(II) oxide decomposes into itselements, mercury and oxygen.Liquid mercury reacts with oxygento re-form mercury(II) oxide.Together these reactions represent a reversible chemical process.

combine to form mercury(II) oxide at the same rate that mercury(II)oxide decomposes into mercury and oxygen.The amounts of mercury(II)oxide, mercury, and oxygen can then be expected to remain constant aslong as these conditions persist. At this point, a state of dynamic equilib-rium has been reached between the two chemical reactions. Both reac-tions continue, but there is no net change in the composition of thesystem. A reversible chemical reaction is in chemical equilibrium whenthe rate of its forward reaction equals the rate of its reverse reaction andthe concentrations of its products and reactants remain unchanged. Thechemical equation for the reaction at equilibrium is written using doublearrows to indicate the overall reversibility of the reaction.

2HgO(s) ⎯→←⎯ 2Hg(l) + O2(g)

Equilibrium, a Dynamic State

Many chemical reactions are reversible under ordinary conditions oftemperature and concentration. They will reach a state of equilibriumunless at least one of the substances involved escapes or is removedfrom the reaction system. In some cases, however, the forward reactionis so predominant that essentially all reactants will react to form prod-ucts. Here, the products of the forward reaction are favored, meaningthat at equilibrium there is a much higher concentration of productsthan of reactants. Hence, we can say that the equilibrium “lies to theright,” because products predominate, and products conventionally arewritten on the right-hand side of a chemical equation. An example ofsuch a system is the formation of sulfur trioxide from sulfur dioxide and oxygen.

2SO2(g) + O2(g) ⎯→←⎯ 2SO3(g)

Notice that the equation is written showing an inequality of the twoarrow lengths. The forward reaction is represented by the longer arrowto imply that the product is favored in this reaction.

In other cases, the forward reaction is barely under way when therate of the reverse reaction becomes equal to that of the forward reac-tion, and equilibrium is established. In these cases, the amounts of reac-tants remain high and the amounts of products are low. Here we saythat the equilibrium “lies to the left,” because the reactants are the pre-dominant species.An example of such a system is the acid-base reactionbetween carbonic acid and water.

H2CO3(aq) + H2O(l) →←⎯ H3O+(aq) + HCO3−(aq)

In still other cases, both forward and reverse reactions occur to near-ly the same extent before chemical equilibrium is established. Neitherreaction is favored, and considerable concentrations of both reactants

C H A P T E R 1 8590

and products are present at equilibrium. An example is the dissociationof sulfurous acid in water.

H2SO3(aq) + H2O(l) ⎯→←⎯ H3O+(aq) + HSO3−(aq)

Chemical reactions ordinarily are used to convert available reactantsinto more desirable products. Chemists try to convert as much of thesereactants as possible into products. The extent to which reactants areconverted to products is indicated by the numerical value of the equi-librium constant.

The Equilibrium Expression

Suppose two substances, A and B, react to form products C and D. Inturn, C and D react to produce A and B. Under appropriate conditions,equilibrium occurs for this reversible reaction. This hypothetical equi-librium reaction is described by the following general equation.

nA + mB ⎯→←⎯ xC + yD

Initially, the concentrations of C and D are zero and those of A andB are maximum. Figure 2 shows that over time the rate of the forwardreaction decreases as A and B are used up. Meanwhile, the rate of thereverse reaction increases as C and D are formed.When these two reac-tion rates become equal, equilibrium is established. The individual con-centrations of A, B, C, and D undergo no further change if conditionsremain the same.

After equilibrium is attained, the concentrations of products andreactants remain constant, so a ratio of their concentrations should also


Time

Rea

ctio

n r

ate

Rate Comparison for the Reaction System A + B ⎯→ C + D

t0 t1

A + B ⎯→ C + D(forward reaction)

C + D ⎯→ A + B(reverse reaction)

Equilibrium

(forward rate = reverse rate)

←⎯

FIGURE 2 Shown are reaction rates for the hypotheticalequilibrium reaction system A + B ⎯→←⎯ C + D. From the timeA and B are mixed together at t0,the rate of the forward reactiondeclines and the rate of the reversereaction increases until both forwardand reverse reaction rates are equalat t1, when the equilibrium conditionbegins.

remain constant. The ratio of the mathematical product [C]x × [D]y tothe mathematical product [A]n × [B]m for this reaction has a definitevalue at a given temperature. It is the equilibrium constant of the reac-tion and is designated by the letter K. The following equation describesthe equilibrium constant for the hypothetical equilibrium system.The brackets ([ ]) indicate the concentration of each substance asexpressed in mol/L. The superscripts are the coefficients of each sub-stance in the balanced chemical equation.

K =

The concentrations of substances on the right side of the chemicalequation appear in the numerator of the ratio, with each concentra-tion raised to a power equal to the coefficient of that substance in thebalanced chemical equation. These substances are the products of theforward reaction. The concentrations of substances on the left side ofthe chemical equation are in the denominator of the ratio, with eachconcentration raised to a power equal to the coefficient of that sub-stance in the balanced chemical equation. These substances are thereactants of the forward reaction. The constant K is independent ofthe initial concentrations. It is, however, dependent on the tempera-ture of the system.

The Equilibrium ConstantThe numerical value of K for a particular equilibrium system isobtained experimentally. The chemist must analyze the equilibriummixture and determine the concentrations of all substances. The valueof K for a given equilibrium reaction at a given temperature shows theextent to which the reactants are converted into the products of thereaction. If the value of K is small, the forward reaction occurs onlyvery slightly before equilibrium is established, and the reactants arefavored. A large value of K indicates an equilibrium in which the orig-inal reactants are largely converted to products. Only the concentra-tions of substances that can actually change are included in K. Thismeans that pure solids and liquids are omitted because their concen-trations cannot change.

In general, then, the equilibrium constant, K, is the ratio of the math-ematical product of the concentrations of substances formed at equilib-rium to the mathematical product of the concentrations of reactingsubstances. Each concentration is raised to a power equal to the coeffi-cient of that substance in the chemical equation. The equation for K issometimes referred to as the chemical equilibrium expression.

The H2, I2, HI Equilibrium SystemConsider the reaction between H2 and I2 vapor in a sealed flask at anelevated temperature. The rate of reaction can be followed by observ-ing the rate at which the violet color of the iodine vapor diminishes, asshown in Figure 3. If colorless H2 gas is present in excess, we might

[C]x[D]y

[A]n[B]m

C H A P T E R 1 8592

expect that the reaction would continue until all of the I2 is used up.Theviolet color of the tube would decrease in intensity until all of the iodinereacts. At that time, the tube would be colorless, because both HI andthe excess H2 are colorless gases.

In actuality, the color fades to a constant intensity but does not dis-appear completely because the reaction is reversible. Hydrogen iodidedecomposes to re-form hydrogen and iodine. The rate of this reversereaction increases as the concentration of hydrogen iodide increases.The rate of the forward reaction decreases accordingly. The concentra-tions of hydrogen and iodine decrease as they react. As the rates of theopposing reactions become equal, equilibrium is established. The con-stant color achieved indicates that equilibrium exists among hydrogen,iodine, and hydrogen iodide. The net chemical equation for the reactionsystem at equilibrium follows.

H2(g) + I2(g) ⎯→←⎯ 2HI(g)

From this chemical equation, the following chemical equilibrium expres-sion can be written. The concentration of HI is raised to the power of 2because the coefficient of HI in the balanced chemical equation is 2.

K =

Chemists have carefully measured the concentrations of H2, I2, andHI in equilibrium mixtures at various temperatures. In some experi-ments, the flasks were filled with hydrogen iodide at known pressure.The flasks were held at fixed temperatures until equilibrium was estab-lished. In other experiments, hydrogen and iodine were the original sub-stances. Experimental data, together with the calculated values for K,are listed in Table 1. Experiments 1 and 2 began with hydrogen iodide.Experiments 3 and 4 began with hydrogen and iodine. Note the closeagreement obtained for the numerical values of the equilibrium con-stant in all cases.

At 425°C, the equilibrium constant for this equilibrium reaction sys-tem has the average value of 54.34. This value for K is constant for any system of H2, I2, and HI at equilibrium at this temperature. If the

[HI]2

[H2][I2]


Experiment [H2] [I2] [HI]K =

1 0.4953 × 10−3 0.4953 × 10−3 3.655 × 10−3 54.46

2 1.141 × 10−3 1.141 × 10−3 8.410 × 10−3 54.33

3 3.560 × 10−3 1.250 × 10−3 15.59 × 10−3 54.62

4 2.252 × 10−3 2.336 × 10−3 16.85 × 10−3 53.97

[HI]2

[H2][I2]

TABLE 1 Typical Equilibrium Concentrations of H2, I2, and HI in mol/L at 425°C

(a)

(b)

(c)

FIGURE 3 Hydrogen iodide gas isproduced from gaseous hydrogenand iodine. The violet color of iodinegas (a) becomes fainter as the reac-tion consumes the iodine (b). Theviolet does not disappear but reach-es a constant intensity when thereaction reaches equilibrium (c).

calculation for K yields a different result, there must be a reason. Eitherthe H2, I2, and HI system has not reached equilibrium or the system isnot at 425°C.

The balanced chemical equation for an equilibrium system is neces-sary to write the expression for the equilibrium constant. The data inTable 1 show that the validity of this expression is confirmed when theactual values of the equilibrium concentrations of reactants and prod-ucts are determined experimentally.The values of K are calculated fromthese concentrations. No information concerning the kinetics of thereacting systems is required.

Once the value of the equilibrium constant is known, the equilibrium-constant expression can be used to calculate concentrations of reactantsor products at equilibrium. Suppose an equilibrium system at 425°C isfound to contain 0.015 mol/L each of H2 and I2. To find the concentra-tion of HI in this system, rearrange the chemical equilibrium expressionas shown in the two equations that follow.

K =

[HI] = �K�[H�2]�[I�2]�

Using the known K value and the given concentrations for H2 and I2,solve the equation for [HI].

[HI] = �0.�01�5�×� 0�.0�15� ×� 5�4.�34�[HI] = 0.11 mol/L

[HI]2

[H2][I2]

C H A P T E R 1 8594

Given: [N2] = 6.4 × 10−3 mol/L[O2] = 1.7 × 10−3 mol/L[NO] = 1.1 × 10−5 mol/L

Unknown: K

The balanced chemical equation is N2(g) + O2(g) ⎯→←⎯ 2NO(g).

The chemical equilibrium expression is K = .

Substitute the given values for the concentrations into the equilibrium expression.

K = = 1.1 × 10−5(1.1 × 10−5 mol/L)2

(6.4 × 10−3 mol/L)(1.7 × 10−3 mol/L)

[NO]2

[N2][O2]

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

SAMPLE PROBLEM A

An equilibrium mixture of N2, O2, and NO gases at 1500 K is determined to consist of 6.4 � 10–3 mol/Lof N2, 1.7 � 10–3 mol/L of O2, and 1.1 � 10–5 mol/L of NO. What is the equilibrium constant for the sys-tem at this temperature?



Go to go.hrw.com formore practice problemsthat ask you to calculateequilibrium constants.

Keyword: HC6EQUX

The value of K is small, which is consistent with more N2 and O2 being present at equilibri-um than NO. The answer has the correct number of significant figures and is close to an estimated value of

8 × 10−6, calculated as .(1 × 10−5)2

(6 × 10−3)(2 × 10−3)

4 EVALUATE

1. At equilibrium a mixture of N2, H2, and NH3 gas at 500°C is deter-mined to consist of 0.602 mol/L of N2, 0.420 mol/L of H2, and 0.113 mol/L of NH3. What is the equilibrium constant for the reac-tion N2(g) + 3H2(g) ⎯→←⎯ 2NH3(g) at this temperature?

2. The reaction AB2C(g) ⎯→←⎯ B2(g) + AC(g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.084 mol of AB2C,0.035 mol of B2, and 0.059 mol of AC were detected. What is theequilibrium constant at this temperature for this system? (Don’tforget to convert amounts to concentrations.)

3. A reaction between gaseous sulfur dioxide and oxygen gas to pro-duce gaseous sulfur trioxide takes place at 600°C. At that temper-ature, the concentration of SO2 is found to be 1.50 mol/L, theconcentration of O2 is 1.25 mol/L, and the concentration of SO3 is3.50 mol/L. Using the balanced chemical equation, calculate theequilibrium constant for this system.


1. What is meant by chemical equilibrium?

2. What is an equilibrium constant?

3. How does the value of an equilibrium constantrelate to the relative quantities of reactants andproducts at equilibrium?

4. What is meant by a chemical equilibriumexpression?

5. Hydrochloric acid, HCl, is a strong acid that disso-ciates completely in water to form H3O+ and Cl−.Would you expect the value of K for the reactionHCl(aq ) + H2O(l ) ⎯→←⎯ H3O+(aq ) + Cl−(aq ) to be1 × 10−2, 1 × 10−5, or “very large”? Justify youranswer.

6. Write the chemical equilibrium expression for thereaction 4HCl(g ) + O2(g ) ⎯→←⎯ 2Cl2(g ) + 2H2O(g ).

7. At equilibrium at 2500 K, [HCl] = 0.0625 mol/Land [H2] = [Cl2] = 0.00450 mol/L for the reactionH2(g ) + Cl2(g ) ⎯→←⎯ 2HCl(g ). Find the value of K.

8. An equilibrium mixture at 425°C is found to con-sist of 1.83 × 10−3 mol/L of H2, 3.13 × 10−3

mol/L of I2, and 1.77 × 10−2 mol/L of HI.Calculate the equilibrium constant, K, for thereaction H2(g ) + I2(g ) ⎯→←⎯ 2HI(g ).

9. For the reaction H2(g ) + I2(g ) ⎯→←⎯ 2HI(g ) at 425°C, calculate [HI], given [H2] = [I2] =4.79 × 10−4 mol/L and K = 54.3.

Critical Thinking

10. INFERRING RELATIONSHIPS Use the data fromExperiment 1 in Table 1 to calculate the valueof K for the reaction 2HI(g ) ⎯→←⎯ H2(g ) + I2(g ).Do you see a relationship between the value youobtained and the value in the table?

SECTION REVIEW

C H A P T E R 1 8596

Fixing the Nitrogen ProblemHistorical PerspectiveEach year, the chemical industry synthesizes tons ofnitrogenous fertilizers, increasing agricultural productionaround the globe. But prior to 1915, humans had to relysolely on natural resources for fertilizer, and the dwindlingsupply of these materials caused widespread fear of worldstarvation. A crisis was averted, however, through the dis-covery of an answer to the “nitrogen problem,” a termused at the time to describe the shortage of useful nitro-gen despite its abundance in the atmosphere.

The Malthusian ThreatIn 1798, Thomas Malthus published his famous “Essay onPopulation,” a report predicting that the world’s food sup-plies could not keep up with the growing human popula-tion and that famine, death, and misery were inevitable.

Animals

Decomposition

Plants

Ammonia

Nitrogen gas

Nitrogen-fixing bacteriaon plant roots

Nitrogen is released when livingthings die and also from animalwastes and plant material. Somebacteria are able to break the bondholding the nitrogen moleculetogether, freeing the nitrogenatoms to combine with hydrogen toform ammonia. Plants can absorbthe nitrogen in this form from thesoil. Animals then benefit from thenitrogen by eating the plants.

Malthus’s warning seemed to be echoed in the 1840s bythe great Irish potato famine. In fact, the rest of Europelikely would have suffered serious food shortages as wellhad crop yields per acre not been increased through theuse of fertilizers containing nitrogen.

Few living things can utilize the gas that forms 78% ofthe atmosphere; they need nitrogen that has been com-bined with other elements, or “fixed,” to survive.

But soil often lacks sufficient amounts of the organismsthat fix nitrogen for plants, so fertilizers containing usablenitrogen compounds are added. In 1898, two-thirds of theworld’s supply of these compounds came from Chile, wherebeds of sodium nitrate, or Chile saltpeter, were abundant.But, as the chemist William Crookes emphasized in hisspeech to the British Association that year, these reserves


were limited; it was up to his colleagues to dis-cover alternatives and prevent Malthus’s direforecast from coming true.

The Haber-Nernst ControversyAs early as the 1890s, chemists had shown thatammonia, a practical source of fixed nitrogen,could be synthesized at high temperatures andat atmospheric pressure from elemental hydro-gen and nitrogen. The problem was that the endproduct was present in such minute amountsthat the process was not industrially practical.

In 1904, the German chemist Fritz Haberseemed to confirm this assessment. He tried react-ing hydrogen and nitrogen at temperatures of upto 1020°C using pure iron as well as other metalsas a catalyst. He found that the amount of ammo-nia was a mere 0.005% to 0.012% at equilibrium.

Haber had apparently closed the door on the synthesisof ammonia from its elements. But in 1906, WaltherNernst, using his new heat theorem, calculated the reac-tion’s theoretical ammonia concentration at equilibria cor-responding to several pressures. He found that his value atatmospheric pressure disagreed significantly with Haber’s,and he publicly challenged Haber’s values.

Haber was convinced that he was right. He ran the reac-tion at increased pressure to attain an amount of ammo-nia that could be measured more accurately.

Haber and his assistants confirmed their original find-ings, and Nernst later conceded a mathematical error. Butmore important, the new round of experiments indicatedthat a reasonable amount of ammonia might be attainedat pressures of 200 atm (20,000 kPa) using a uranium orosmium catalyst.

Scaling UpLarge-scale equipment that could withstand such highpressures was unheard of at the time, and osmium anduranium were far too scarce to be cost-effective for indus-try. Nevertheless, in 1909, the German firm BASF boughtthe rights to Haber’s findings and put its gifted chemicalengineer Karl Bosch in charge of creating an industrial-scale system that would make the process profitable.

After nearly five years, Bosch and the company’s chiefchemist, Alwin Mittasch, succeeded in developing a suit-able reactor that could handle the reaction’s high pres-

sures. They also discovered that a catalyst of iron contain-ing small amounts of impurities was an effective replace-ment for the rare metals used by Haber.

An Eerie EpilogueBy September 1913, BASF was producing 20 metric tons ofammonia a day using the Haber-Bosch process. Eventually,enough ammonia was produced by the chemical industryto free Germany and the world of dependence on Chilesaltpeter for fertilizer. Chemists had thwarted theMalthusian threat. Yet, the victory proved bittersweet; thenew ammonia synthesis also became the basis of the pro-duction of nitric acid, used to make many of the explosivesemployed in the wars that rocked Europe and the rest ofthe globe in the first half of the twentieth century.

1. What is the major use for ammonia?

2. What did Haber find when he tried to synthesizeammonia at increased pressure?

Questions

Today, ammonia is produced on an industrial scale in plantslike this one.

www.scilinks.orgTopic: NitrogenCode: HC61035

C H A P T E R 1 8598

SECTION 2 Shifting Equilibrium

I n systems that have attained chemical equilibrium, the relativeamounts of reactants and products stay the same. But changes in pres-sure, concentration, or temperature can alter the equilibrium positionand thereby change the relative amounts of reactants and products. Byshifting an equilibrium in the desired direction, chemists can oftenimprove the yield of the product they are seeking.

Predicting the Direction of Shift

In 1888, the French chemist Henri Louis Le Châtelier developed a prin-ciple that provides a means of predicting the influence of stress factors onequilibrium systems. Le Châtelier’s principle states that if a system atequilibrium is subjected to a stress, the equilibrium is shifted in the direc-tion that tends to relieve the stress. This principle is true for all dynamicequilibria, chemical as well as physical. Changes in pressure, concentra-tion, and temperature illustrate Le Châtelier’s principle.

Changes in PressureA change in pressure affects only equilibrium systems in which gases are involved. For changes in pressure to affect the system, the total number of moles of gas on the left side of the equation must be differentfrom the total number of moles of gas on the right side of the equation.

Let us consider the Haber process for the synthesis of ammonia. Notethat there is a total of four molecules of gas on the reactant side of theequation and two molecules of gas on the product side of the equation.

N2(g) + 3H2(g) ⎯→←⎯ 2NH3(g)

First, consider an increase in pressure as the applied stress. Can the sys-tem shift in a way that reduces the stress? Yes. An increase in pressurecauses increases in the concentrations of all species. The system canreduce the number of molecules, and hence the total pressure, by shift-ing the equilibrium to the right. For each four molecules of reactants,nitrogen and hydrogen, there are two molecules of product, ammonia.By producing more NH3, and using up N2 and H2, the system can reducethe total number of molecules. This leads to a decrease in pressure.Although the new equilibrium pressure is still higher than before, it isnot as high as the pressure caused by the initial stress.

An increase in pressure on confined gases causes an increase in theconcentrations of these gases. Even though changes in pressure may

OBJECTIVES

Discuss the factors that disturb equilibrium.

Discuss conditions under whichreactions go to completion.

Describe the common-ioneffect.

www.scilinks.orgTopic: Haber ProcessCode: HC60704

shift the equilibrium position, they do not affect the value of the equi-librium constant.

Ammonia produced in the Haber process is continuously removed bycondensing it to liquid ammonia.This condensation removes most of theproduct from the gas phase in which the reaction occurs. The resultingdecrease in the partial pressure of NH3 gas in the reaction vessel is astress and is the same as a decrease in product concentration, whichshifts the equilibrium to the right.

The introduction of an inert gas, such as helium, into the reaction ves-sel for the synthesis of ammonia increases the total pressure in the vessel.But it does not change the partial pressures of the reaction gases present.Therefore, increasing pressure by adding a gas that is not a reactant ora product cannot affect the equilibrium position of the reaction system.

Changes in ConcentrationAn increase in the concentration of a reactant is a stress on the equilib-rium system. Consider the following hypothetical reaction.

A + B ⎯→←⎯ C + D

An increase in the concentration of A creates a stress. To relieve thestress, some of the added A reacts with B to form products C and D. Theequilibrium is reestablished with a higher concentration of A than beforethe addition but a lower concentration of B. Figure 4 illustrates the effecton a system in equilibrium produced by increasing the concentration of areactant. Similarly, an increase in the concentration of B drives the reac-tion to the right. An increase in the concentration of either C or D shiftsthe equilibrium to the left.A decrease in the concentration of C or D hasthe same effect on the position of the equilibrium as does an increase inthe concentration of A or B; the equilibrium shifts to the right.

Changes in concentration have no effect on the value of the equilib-rium constant. Although concentrations of both reactants and productsdo change, the new concentrations give the same value of the equilibri-um constant when equilibrium is reestablished.

Many chemical processes involve heterogeneous reactions in whichreactants or products are in different phases. The concentrations of puresolids and liquids do not change, and by convention are not written in theequilibrium expression. Also, when a solvent such as water, in a systeminvolving acids and bases, is in an equilibrium equation, it is not includ-ed in the equilibrium expression. In Chapter 15, the expression for Kwused this convention and the concentration of water is not included inthe expression. The reaction representing the self-ionization of water is

2H2O(l) ⎯→←⎯ H3O+(aq) + OH−(aq)

and the equation for Kw is Kw = [H3O+][OH−].The following equation describes the equilibrium system established

by the decomposition of solid calcium carbonate.

CaCO3(s) ⎯→←⎯ CaO(s) + CO2(g)


(a) at equilibrium

(b) stressed (not at equilibrium)

(c) at new equilibrium

N2(g) + 3H2(g) ⎯→←⎯ 2NH3(g)

N2

H2

NH3

N2 added

FIGURE 4 (a) N2, H2, and NH3are in equilibrium within a closedsystem. (b) Addition of more N2causes a stress on the initial equilib-rium. (c) The new equilibrium posi-tion for this system has a higherconcentration of N2, a lower concen-tration of H2, and a higher concen-tration of NH3 than initially.

The products are a solid and a gas. Because both CaCO3 and CaO aresolids, they are not in the equilibrium constant expression. This leads tothe following expression for the equilibrium constant.

K = [CO2]

Carbon dioxide is the only substance in the system that appears in theequilibrium expression. Because the total number of moles of gas onthe left side of the equation is different from the total number of moleson the right side of the equation, pressure changes will affect the equi-librium. High pressure favors the reverse reaction, which causes CO2molecules to react with the solid CaO to form solid CaCO3. Low pres-sure favors the formation of CO2 from the decomposition of CaCO3.Because both CaO and CaCO3 are solids, changing their amounts willnot change the equilibrium concentration of CO2.

Changes in TemperatureReversible reactions are exothermic in one direction and endothermicin the other. Remember, equilibrium constants are for a given temper-ature because changing the temperature changes the relative amountsof reactants and products.

Increasing the temperature is, in effect, the addition of enery in theform of heat. According to Le Châtelier’s principle, the stress of theadded heat will be lessened by shifting the equilibrium in the directionthat removes heat (lowers the temperature). This means that energymust be absorbed so the reaction that is endothermic occurs until a newequilibrium is established. Likewise, the removal of energy as a result oflowering the temperature causes the exothermic reaction to take place.

The synthesis of ammonia by the Haber process is exothermic, as indi-cated by the energy as heat shown on the product side of the equation.

N2(g) + 3H2(g) ⎯→←⎯ 2NH3(g) + 92 kJ

A high temperature favors the decomposition of ammonia, theendothermic reaction. But at low temperatures, the forward reaction istoo slow to be commercially useful. The temperature used represents acompromise between kinetic and equilibrium requirements. It is highenough that equilibrium is established rapidly but low enough that theequilibrium concentration of ammonia is significant. Moderate temper-ature (about 500°C) and very high pressure (700–1000 atm) produce asatisfactory yield of ammonia.

The production of colorless dinitrogen tetroxide gas, N2O4, from darkred-brown NO2 gas is also an exothermic reaction. Figure 5 shows howtemperature affects the equilibrium of this system. Figure 5b shows theNO2/N2O4 equilibrium mixture at 25°C.When the temperature of the sys-tem is lowered to 0°C, the system experiences a stress (removal of ener-gy as heat). To counteract this stress, the system shifts to the right, or inthe direction of the exothermic reaction. This shift increases the amountof colorless N2O4 gas and decreases the amount of brown NO2 gas, asshown in Figure 5a. Because more N2O4 is present, K is increased. Whenthe system is heated to 100°C, the added energy is the stress, and the equi-

C H A P T E R 1 8600

librium shifts to the left, or in the direction of the endothermic reaction.This shift decreases the amount of colorless N2O4 gas and increases theamount of brown NO2 gas, as shown in Figure 5c. Because less N2O4 gasis present, K is decreased. The change in temperature changes the valueof K. For any system in which the forward reaction is an exothermic reac-tion, increasing the temperature decreases the value of K.

For an endothermic reaction, such as the decomposition of calciumcarbonate, energy as heat shows up on the reactant side of the equation.

556 kJ + CaCO3(s) ⎯→←⎯ CaO(s) + CO2(g)

An increase in temperature caused by adding energy to the system causes the value of K to increase and the equilibrium to shift to the right.

Catalysts speed up the rate of reactions. So what happens to equilib-rium concentrations if a catalyst is present? Nothing! When a catalyst isadded to an equilibrium system, it speeds up both the forward andreverse reactions. The equilibrium concentrations are achieved faster,but the concentrations and K remain the same.

Reactions That Go to Completion

Some reactions involving compounds formed by the chemical interac-tion of ions in solutions appear to go to completion in the sense that theions are almost completely removed from solution. The extent to whichreacting ions are removed from solution depends on the solubility ofthe compound formed and, if the compound is soluble, on the degree ofionization. Thus, a product that escapes as a gas, precipitates as a solid,or is only slightly ionized effectively removes from solution the bulk ofthe reacting ions that compose it. Consider some specific examples ofsituations in which such ionic reactions go to completion.


0°C 25°C 100°CVery light brown Medium brown Dark brown

2NO2(g) ⎯→←⎯ N2O4(g) + energy as heat

(a) (b) (c)FIGURE 5 Different temperaturescan cause an equilibrium system toshift and seek a new equilibriumposition.

(brown) (colorless)

Formation of a GasReactions that form a gas as a product are one example of reactionsthat go to completion. When a strong acid is added to an aqueous solu-tion of baking soda, or sodium bicarbonate, carbon dioxide is formed.The net ionic equation shows that ions are removed.

H3O+ + HCO−3 ⎯→ 2H2O(l) + CO2(g)

This reaction goes practically to completion because one of the prod-ucts, CO2, escapes as a gas if the container is open to the air.

Formation of a PrecipitateWhen solutions of sodium chloride and silver nitrate are mixed, a whiteprecipitate of silver chloride immediately forms, as shown in Figure 6.The overall ionic equation for this reaction follows.

If chemically equivalent amounts of the two solutes are mixed, only Na+

ions and NO3− ions remain in solution in appreciable quantities. Almost

all of the Ag+ ions and Cl− ions combine and separate from the solutionas a precipitate of AgCl. The reason is that AgCl is only very slightlysoluble in water. The solution is now a saturated solution of AgCl. Thereaction thus effectively goes to completion because an essentiallyinsoluble product is formed.

C H A P T E R 1 8602

Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) ⎯→ Na+(aq) + NO3

−(aq) + AgCl(s)

Cl−

H2ONa+

Cl−

AgCl

Na+

NO3− H2O

Ag+

NO3−

H2O

Ag+

FIGURE 6 When a clear sodiumchloride solution is combined with a clear solution of silver nitrate, aninsoluble white precipitate of silverchloride is formed.

Formation of a Slightly Ionized ProductNeutralization reactions between H3O+ ions from aqueous acids andOH− ions from aqueous bases result in the formation of water molecules,which are only slightly ionized. A reaction between HCl and NaOHillustrates this process.Aqueous HCl supplies H3O+ ions and Cl− ions tothe solution, and aqueous NaOH supplies Na+ ions and OH− ions, asshown in the following overall ionic equation.

Neglecting the spectator ions, the net ionic equation is as follows.

H3O+(aq) + OH−(aq) ⎯→ 2H2O(l)

Because it is only slightly ionized, the water exists almost entirely asmolecules. Thus, hydronium ions and hydroxide ions are almost entirelyremoved from the solution. The reaction effectively goes to completionbecause the product is only slightly ionized.

Common-Ion Effect

An equilibrium reaction may be driven in the desired direction byapplying Le Châtelier’s principle. Suppose hydrogen chloride gas isbubbled into a saturated solution of sodium chloride. Hydrogen chlo-ride is extremely soluble in water, and it is completely ionized.

HCl(g) + H2O(l) ⎯→ H3O+(aq) + Cl−(aq)

The equilibrium for a saturated solution of sodium chloride is describedby the following equation.

NaCl(s) ⎯→←⎯ Na+(aq) + Cl−(aq)

As the hydrogen chloride dissolves in sufficient quantity, it increases theconcentration of Cl− ions in the solution, which is a stress on the equi-librium system. The system can compensate, according to Le Châtelier’sprinciple, by combining some of the added Cl− ions with an equivalentamount of Na+ ions. This causes some solid NaCl to precipitate out,relieving the stress of added chloride.The new equilibrium has a greaterconcentration of Cl− ions but a decreased concentration of Na+ ions.However, the product of [Na+] and [Cl−] still has the same value asbefore. This phenomenon, in which the addition of an ion common totwo solutes brings about precipitation or reduced ionization, is an exam-ple of the common-ion effect.

The common-ion effect is also observed when one ion species of aweak electrolyte is added in excess to a solution.Acetic acid, CH3COOH,


H3O+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) ⎯→ Na+(aq) + Cl−(aq) + 2H2O(l)

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Keyword: HC6EQUX

is such an electrolyte. A 0.1 M CH3COOH solution is only about 1.3%ionized as hydronium ions and acetate ions, CH3COO−. The ionic equi-librium is shown by the following equation.

CH3COOH(aq) + H2O(l) ⎯→←⎯ H3O+(aq) + CH3COO−(aq)

Addition of sodium acetate, NaCH3COO (an ionic salt that is soluble inwater), to a solution containing acetic acid increases the acetate ion con-centration. The equilibrium then shifts in the direction that uses up someof the added acetate ions in accordance with Le Châtelier’s principle.More molecules of acetic acid are formed, and the concentration ofhydronium ions is reduced. In general, the addition of a salt with an ionin common with the weak electrolyte reduces the ionization of the elec-trolyte. Figure 7 shows a 0.25 M CH3COOH solution on the left that has

a pH of about 2.7.Mixing that with the0.10 M NaCH3COOsolution in the centerproduces the solutionon the right, which hasa pH of about 4.5, indi-cating lower [H3O

+] andthus lowered acetic acidionization.

C H A P T E R 1 8604

FIGURE 7 The solution ofCH3COOH on the left is combinedwith the solution of NaCH3COO inthe center. Both contain the com-mon ion, CH3COO−. They producethe solution on the right, which is only slightly acidic due to thedecreased ionization of the acid. Thecolors of the solutions are due to theaddition of an acid-base indicator.

1. Name three ways the chemical equilibrium can bedisturbed.

2. Describe three situations in which ionic reactionsgo to completion.

3. Describe the common-ion effect.

4. Identify the common ion in each of the followingsituations.

a. 5 g of NaCl is added to a 2.0 M solution of HCl

b. 50 mL of 1.0 M NaCH3COO is added to 1.0 MCH3COOH

c. 1 g of NH4Cl is added to 100 mL of aqueous NH3

5. Predict the effect that decreasing pressure wouldhave on each of the following reaction systems atequilibrium.

a. H2(g ) + Cl2(g) ⎯→←⎯ 2HCl(g )

b. NH4Cl(s ) ⎯→←⎯ NH3(g ) + HCl(g )

c. 2H2O2(aq ) ⎯→←⎯ 2H2O(l ) + O2(g )

d. 3O2(g ) ⎯→←⎯ 2O3(g )

Critical Thinking

6. PREDICTING OUTCOMES Carbon dioxide and waterreact to form bicarbonate ion and hydronium ion.Hyperventilation (rapid breathing) causes more car-bon dioxide to be exhaled than normal. How willhyperventilation affect the pH of blood? Explain.

SECTION REVIEW


SECTION 3

OBJECTIVES

Explain the concept of acid ionization constants,and write acid ionization equilibrium expressions.

Review the ionizationconstant of water.

Explain buffering.

Compare cation and anionhydrolysis.

Equilibria of Acids,Bases, and Salts

Ionization Constant of a Weak Acid

About 1.3% of the solute molecules in a 0.1 M acetic acid solution areionized at room temperature. The remaining 98.7% of the acetic acidmolecules, CH3COOH, remain nonionized. Thus, the solution containsthree species of particles in equilibrium: CH3COOH molecules, H3O+

ions, and acetate ions, CH3COO−. From the equilibrium equation forthe ionization of acetic acid, the equilibrium constant equation can bewritten.

CH3COOH + H2O ⎯→←⎯ H3O+ + CH3COO−

K =

Notice that the concentration of water is not included in the equilib-rium expression. The reason is that water is the solvent, and water mol-ecules greatly exceed the number of acetic acid molecules. Withoutintroducing a measurable error, one can assume that the molar concen-tration of H2O molecules remains constant in such a solution. Thus,because both K and [H2O] are constant, the product K[H2O] is constant.

K [H2O] =

The left side of the equation can be simplified by setting K[H2O] = Ka.

Ka =

The term Ka is called the acid ionization constant. The acid ionizationconstant, Ka, like the equilibrium constant, K, is constant for a specifiedtemperature but has a new value for each new temperature.

The acid ionization constant for a weak acid represents a small value.To determine the numerical value of the ionization constant for aceticacid at a specific temperature, the equilibrium concentrations of H3O+

ions, CH3COO− ions, and CH3COOH molecules must be known. Theionization of a molecule of CH3COOH in water yields one H3O+ ionand one CH3COO− ion. These concentrations can, therefore, be foundexperimentally by measuring the pH of the solution.

[H3O+][CH3COO−]

[CH3COOH]

[H3O+][CH3COO−]

[CH3COOH]

[H3O+][CH3COO−]

[CH3COOH][H2O]

Ionization data and constants for some dilute acetic acid solutions at25°C are given in Table 2. Notice that the numerical value of Ka isalmost identical for each solution molarity shown. The numerical valueof Ka for CH3COOH at 25°C can be determined by substituting numer-ical values for concentration into the equilibrium equation.

Ka =

At constant temperature,an increase in the concentration of CH3COO−

ions through the addition of sodium acetate, NaCH3COO, disturbs theequilibrium, as predicted by Le Châtelier’s principle. This disturbancecauses a decrease in [H3O

+] and an increase in [CH3COOH]. Eventually,the equilibrium is reestablished with the same value of Ka . But there is ahigher concentration of nonionized acetic acid molecules and a lower con-centration of H3O

+ ions than before the extra CH3COO− was added.Changes in the hydronium ion concentration affect pH. In this example,the reduction in [H3O

+] means an increase in the pH of the solution.

Buffers

The solution just described contains both a weak acid, CH3COOH, anda salt of the weak acid, NaCH3COO. The solution can react with eitheran acid or a base. When small amounts of acids or bases are added, thepH of the solution remains nearly constant.The weak acid and the com-mon ion, CH3COO−, act as a “buffer” against significant changes in thepH of the solution. Because it can resist changes in pH, this solution is abuffered solution. Figure 8 shows how a buffered and a nonbufferedsolution react to the addition of an acid.

Suppose a small amount of acid is added to the acetic acid–sodiumacetate solution. Acetate ions react with most of the added hydroniumions to form nonionized acetic acid molecules.

CH3COO−(aq) + H3O+(aq) ⎯→ CH3COOH(aq) + H2O(l)

[H3O+][CH3COO−]

[CH3COOH]

C H A P T E R 1 8606

buffered unbuffered

buffered unbuffered

(a)

(b)

FIGURE 8 (a) The beaker on theleft contains a buffered solution andan indicator with a pH of about 5.The beaker on the right containsmostly water with a trace amount ofacid and an indicator. The pH metershows a pH of 5.00 for this solution.(b) After 5 mL of 0.10 M HCl isadded to both beakers, the beakeron the left does not change color,indicating no substantial change inits pH. However, the beaker on theright undergoes a definite colorchange, and the pH meter shows a pH of 2.17.

Molarity % ionized [H3O+] [CH3COOH] Ka

0.100 1.33 0.00133 0.0987 1.79 × 10−5

0.0500 1.89 0.000945 0.0491 1.82 × 10−5

0.0100 4.17 0.000417 0.00958 1.81 × 10−5

0.00500 5.86 0.000293 0.00471 1.82 × 10−5

0.00100 12.6 0.000126 0.000874 1.82 × 10−5

TABLE 2 Ionization of Acetic Acid

The hydronium ion concentration and the pH of the solution remainpractically unchanged.

Suppose a small amount of a base is added to the original solution.The OH− ions of the base react with and remove hydronium ions to formnonionized water molecules. Acetic acid molecules then ionize andmostly replace the hydronium ions neutralized by the added OH− ions.

CH3COOH(aq) + H2O(l) ⎯→ H3O+(aq) + CH3COO−(aq)

The pH of the solution again remains practically unchanged.A solution of a weak base containing a salt of the base behaves in a

similar manner. The hydroxide ion concentration and the pH of thesolution remain essentially constant with small additions of acids orbases. Suppose a base is added to an aqueous solution of ammonia thatalso contains ammonium chloride. Ammonium ions donate a proton tothe added hydroxide ions to form nonionized water molecules.

NH4+(aq) + OH−(aq) ⎯→ NH3(aq) + H2O(l)

If a small amount of an acid is added to the solution instead, hydrox-ide ions from the solution accept protons from the added hydroniumions to form nonionized water molecules. Ammonia molecules in thesolution then ionize and mostly replace the hydroxide ions neutralizedby added H3O+.

NH3(aq) + H2O(l) ⎯→ NH4+(aq) + OH−(aq)

Buffer action has many important applications in chemistry andphysiology. Human blood is naturally buffered to maintain a pH ofbetween 7.3 and 7.5.This is essential because large changes in pH wouldlead to serious disturbances of normal body functions. Figure 9 showsan example of one of the many medicines buffered to prevent large andpotentially damaging changes in pH.

Ionization Constant of Water

Recall from Chapter 15 that the self-ionization of water is an equilib-rium reaction.

H2O(l) + H2O(l) ⎯→←⎯ H3O+(aq) + OH−(aq)

Equilibrium is established with a very low concentration of H3O+

and OH− ions. The expression for the equilibrium constant, Kw =[H3O+][OH−], is derived from the balanced chemical equation. Thenumerical value of Kw, obtained experimentally, is 1.0 × 10−14 at 25°C.


FIGURE 9 Many consumer prod-ucts are buffered to protect the bodyfrom potentially harmful pH changes.

www.scilinks.orgTopic: BuffersCode: HC60195

Hydrolysis of Salts

Salts are formed during neutralization reactions between acids andbases. When a salt dissolves in water, it produces positive ions (cations)of the base from which it was formed and negative ions (anions) of theacid from which it was formed. Therefore, the solution might be expect-ed to be neutral. The aqueous solutions of some salts, such as NaCl andKNO3, are neutral, having a pH of 7. However, when sodium carbonatedissolves in water, the resulting solution turns red litmus paper blue,indicating a pH greater than 7. Ammonium chloride produces an aque-ous solution that turns blue litmus paper red, indicating a pH less than7. Salts formed from the combination of strong or weak acids and basesare shown in Figure 10.

The variation in pH values can be accounted for by examining theions formed when each of these salts dissociates. If the ions formed arefrom weak acids or bases, they react chemically with the water mole-cules, and the pH of the solution will have a value other than 7. A reac-tion between water molecules and ions of a dissolved salt is hydrolysis. Ifthe anions react with water, the process is anion hydrolysis and resultsin a more basic solution. If the cations react with water molecules, theprocess is cation hydrolysis and results in a more acidic solution.

Anion HydrolysisIn the Brønsted sense, the anion of the salt is the conjugate base of theacid from which it was formed. It is also a proton acceptor. If the acid isweak, its conjugate base (the anion) will be strong enough to remove pro-

C H A P T E R 1 8608

(a) (b) (c) (d)

FIGURE 10 The universal indicator shows that the pH of salt solutionsvaries, depending on the strength of the acid and the base that formed the salt.(a) NaCl is formed from a strong acid and a strong base; the color of the indi-cator shows the pH is neutral. (b) The indicator shows the pH of the sodiumacetate solution is basic. This was formed from a strong base and a weak acid.(c) The strong acid and weak base combination in ammonium chloride pro-duces an acidic solution, as shown by the reddish tint of the indicator. (d) Theweak acid and weak base that form ammonium acetate are of comparablestrength. A solution of ammonium acetate is essentially neutral.

tons from some water molecules, proton donors, to form OH− ions. Anequilibrium is established in which the net effect of the anion hydrolysisis an increase in the hydroxide ion concentration, [OH−], of the solution.

The equilibrium equation for a typical weak acid in water, HA, form-ing hydronium ion and an anion, A−, is as follows.

HA(aq) + H2O(l) ⎯→←⎯ H3O+(aq) + A−(aq)

From this equation, the generalized expression for Ka can be written.Note that as before, water does not appear in the general equilibriumequation.

Ka =

The hydrolysis reaction between water and the anion, A−, that is pro-duced by the ionization of the weak acid, HA, is represented by the gen-eral equilibrium equation that follows.

A−(aq) + H2O(l) ⎯→←⎯ HA(aq) + OH−(aq)

Neutral water has equal concentrations of H3O+ and OH−. Since HA is aweak acid, the anion A− has a strong attraction for protons.Adding A− towater in effect attracts (removes) H3O+ in water to form HA. This caus-es OH− to increase relative to H3O+ as represented in the equationabove.The lower the value of Ka, the stronger the attraction A− will havefor protons and the larger the concentration of OH−. In other words, theweaker the acid, HA, the stronger its conjugate base, A−.

Aqueous solutions of sodium carbonate are strongly basic. The so-dium ions, Na+, in sodium carbonate do not undergo hydrolysis in aque-ous solution, but the carbonate ions, CO3

2−, react as a Brønsted base. ACO3

2− anion acquires a proton from a water molecule to form the weakBrønsted acid, HCO3

−, and the OH− ion.

CO32−(aq) + H2O(l) ⎯→←⎯ HCO3

−(aq) + OH−(aq)

The OH− ion concentration increases until equilibrium is established.Consequently, the H3O+ ion concentration decreases so that the prod-uct [H3O+][OH−] remains equal to the ionization constant, Kw, of waterat the temperature of the solution.Thus, the pH is higher than 7, and thesolution is basic.

Cation HydrolysisIn the Brønsted sense, the cation of the salt is the conjugate acid of thebase from which it was formed. It is also a proton donor. If the base isweak, the cation is an acid strong enough to donate a proton to a watermolecule, a proton acceptor, to form H3O+ ions. An equilibrium isestablished in which the net effect of the cation hydrolysis is an increasein the hydronium ion concentration, [H3O+], of the solution.

[H3O+][A−]

[HA]


Blood BuffersBlood normally has a pH of 7.4. If thepH of blood in a human rises above 7.8or falls below 7.0, this change in pH isusually fatal. The primary buffer in bloodis the carbonic acid–bicarbonate ion sys-tem, CO2(g ) + H2O(l ) ⎯→←⎯ H2CO3(aq ) +HCO3

−(aq ). A condition called respiratoryacidosis causes the pH to drop. It is theresult of hypoventilation, or slowedbreathing, the opposite of hyperventila-tion. Hypoventilation can be caused bycongestive heart failure, pneumonia, ornarcotics. Because breathing is slowed,carbon dioxide accumulates and its concentration rises. This change causes ashift in the equilibrium to the right dueto Le Châtelier’s principle, and the bloodbecomes more acidic because of thehigher concentration of carbonic acid.

The following equilibrium equation for a typical weak base, B, is usedto derive the generalized expression for Kb, the base dissociationconstant.

B(aq) + H2O(l) ⎯→←⎯ BH+(aq) + OH−(aq)

Kb =

The hydrolysis reaction between water and the cation, BH+, producedby the dissociation of the weak base, B, is represented by the generalequilibrium equation that follows.

BH+(aq) + H2O(l) ⎯→←⎯ H3O+(aq) + B(aq)

In the forward reaction, the cation BH+ donates a proton to the watermolecule to form the hydronium ion. Because H3O+ ions are formed,the solution must become more acidic, as shown in the equation above.The extent of H3O+ ion formation depends on the relative strength ofthe base B. The weaker the base, the greater the concentration of H3O+

ions will be. Therefore, the weaker the base, the stronger its conjugateacid.

Ammonium chloride, NH4Cl, dissociates in water to produce NH4+

ions and Cl− ions. Chloride ions are the conjugate base of the strongacid HCl, so they do not hydrolyze in water. Ammonium ions, however,are the conjugate acid of a weak base, NH3. Ammonium ions donateprotons to water molecules. Equilibrium is established with anincreased [H3O+], so the pH is lower than 7.

[BH+][OH−]

[B]

C H A P T E R 1 8610

Neutralization Curve for 100 mL of 0.100 M CH3COOH Titrated with 0.100 M NaOH

mL NaOH added

pH

13

12

11

10

9

8

7

6

5

4

3

2

1

00 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160

1

3

2

Equivalence point

Buffer region

4

FIGURE 11 At point 1 on the titration curve, only acetic acid is present. The pH depends on the weakacid alone. At 2 there is a mixture ofCH3COOH and CH3COO−. AddingNaOH changes the pH slowly. Atpoint 3 all acid has been converted to CH3COO−. This hydrolyzes to produce a slightly basic solution.At 4 the pH is due to the excess OH− that has been added.

Hydrolysis in Acid-Base ReactionsHydrolysis can help explain why the end point of a neutralization reac-tion can occur at a pH other than 7. The hydrolysis properties of saltsare determined by the relative strengths of the acids and bases fromwhich the salts were formed. Salts can be placed in four general cate-gories, depending on their hydrolysis properties: strong acid–strongbase, strong acid–weak base, weak acid–strong base, and weak acid–weak base.

Salts of strong acids and strong bases produce neutral solutionsbecause neither the cation of a strong base nor the anion of a strongacid hydrolyzes appreciably in aqueous solutions. HCl(aq) is a strongacid, and NaOH(aq) is a strong base. Neither the Na+ cation of thestrong base nor the Cl− anion of the strong acid undergoes hydrolysis inwater solutions. Therefore, aqueous solutions of NaCl are neutral.Similarly, KNO3 is the salt of the strong acid HNO3 and the strong baseKOH. Measurements show that the pH of an aqueous KNO3 solution isalways very close to 7.

The aqueous solutions of salts formed from reactions between weakacids and strong bases are basic at the equivalence point, as shown inFigure 11. Anions of the dissolved salt are hydrolyzed by the watermolecules, and the pH of the solution is raised, indicating that thehydroxide-ion concentration has increased. Aqueous solutions of sodi-um acetate, NaCH3COO, are basic.The acetate ions, CH3COO−, under-go hydrolysis because they are the anions of the weak acid acetic acid.The cations of the salt are from a strong base, NaOH, and do nothydrolyze, because NaOH is 100% dissociated.

Figure 12 shows that salts of strong acids and weak bases are acidicat the equivalence point. Cations of the dissolved salt are hydrolyzed in


FIGURE 12 At point 1 on thetitration curve, only aqueous ammo-nia is present. The pH is determinedby the base alone. At 2 there is amixture of NH3 and NH4

+. AddingHCl changes the pH slowly. Atpoint 3 all aqueous ammonia hasbeen converted to NH4

+. At 4 the pH is determined by the excessH3O+ that is being added.

Neutralization Curve for 100 mL of 0.100 M NH3 Titrated with 0.100 M HCl

mL HCl added

pH

13

12

11

10

9

8

7

6

5

4

3

2

1

00 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160

1

3

2

Equivalence point

4

Buffer region

the water solvent, and the pH of the solution is lowered, indicating thatthe hydronium ion concentration has increased. In this case, the cationsof the salt undergo hydrolysis because they are the positive ions from aweak base. The anions of the salt are the negative ions from a strongacid and do not hydrolyze appreciably. Ammonium chloride, NH4Cl, isa salt that produces an acidic solution.

Salts of weak acids and weak bases can produce either acidic, neu-tral, or basic aqueous solutions, depending on the salt dissolved. This isbecause both ions of the dissolved salt are hydrolyzed extensively. Ifboth ions are hydrolyzed equally, the solution remains neutral. The ionsin ammonium acetate, NH4CH3COO, hydrolyze equally, producing aneutral solution, as shown in Figure 10d on page 608.

In salts formed from a weak acid and weak base, the cation and anionboth undergo hydrolysis. For example, when aluminum sulfide is placedin water,Al3+ reacts with OH– forming Al(OH)3, and S2– reacts with H+

forming H2S. The reaction is symbolized by the following chemicalequation.

Al2S3(s) + 6H2O(l) ⎯→ 2Al(OH)3(s) + 3H2S(g)

Since Al(OH)3 is a precipitate and H2S is a gas, both are removed fromsolution.

C H A P T E R 1 8612

1. What is meant by an acid ionization constant?

2. How is an acid ionization equilibrium expressionwritten?

3. What is meant by the term buffered solution?

4. Which of the following combinations of solutionswould form buffers when they are mixed?

a. 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaCl

b. 25 mL of 0.5 M HNO2 and 50 mL of 1.0 MNaNO2

c. 25 mL of 1.0 M HNO2 and 25 mL of 1.0 M NaCl

5. What is meant by the ion product constant forwater? What is the value of this constant?

6. For each of the following reactions, identify eachconjugate acid-base pair.

a. H2CO3 + H2O ⎯→←⎯ HCO3− + H3O+

b. H2O + H2O ⎯→←⎯ H3O+ + OH−

c. H2S + NH3⎯→←⎯ HS− + NH4

+

d. H2PO4− + H2O ⎯→←⎯ H3PO4 + OH−

7. What is hydrolysis? Compare cation and anionhydrolysis.

8. Which of the following ions hydrolyze in aqueoussolution?

a. NO3− d. K+ g. CO3

2−

b. F− e. CH3COO− h. PO43−

c. NH4+ f. SO4

2−

9. Identify the following solutions as acidic, basic,or neutral.

a. 0.5 M KI c. 0.25 M NH4NO3

b. 0.10 M Ba(OH)2 d. 0.05 M K2CO3

10. Identify the acid and base from which each of the following salts was formed.

a. K2CrO4 c. CaF2

b. Ca(CH3COO)2 d. (NH4)2SO4

Critical Thinking

11. RELATING IDEAS Describe how to make a buffersolution using a strong base and one otherreagent.

SECTION REVIEW


SECTION 4

OBJECTIVES

Explain what is meant bysolubility product constants,and calculate their values.

Calculate solubilities usingsolubility product constants.

Carry out calculations to predict whether precipitateswill form when solutions are combined.

Solubility Equilibrium

I onic solids dissolve in water until they are in equilibrium with theirions.An equilibrium expression can be written from the balanced chemi-cal equation of the solid’s dissociation. Concentrations of the ions canbe determined from the balanced chemical equation and solubility data.The ion concentrations can then be used to determine the value of theequilibrium constant. The numerical value for the equilibrium constantcan be used to predict whether precipitation occurs when solutions ofvarious concentrations are combined.

Solubility Product

A saturated solution contains the maximum amount of solute possibleat a given temperature in equilibrium with an undissolved excess of thesubstance. A saturated solution is not necessarily a concentrated solu-tion. The concentration may be high or low, depending on the solubilityof the solute.

A general rule is often used to express solubilities qualitatively. Bythis rule, a substance is said to be soluble if the solubility is greater than1 g per 100 g of water and slightly soluble if less than that. Even sub-stances we have previously referred to as “insoluble” are slightly solu-ble. We will describe the degree of solubility with an equilibriumconstant.

The equilibrium principles developed in this chapter apply to all satu-rated solutions of slightly soluble salts. Silver chloride is an example ofa slightly soluble salt. Its solution reaches saturation when the Ag+ andCl− concentrations are 1.3 × 10−5 M, or about 2 × 10−4 g of AgCl in 100 mL. When mixed, all Ag+ and Cl− ions in excess of this concentra-tion eventually precipitate as AgCl.

Consider the equilibrium system in a saturated solution of silverchloride containing an excess of the solid AgCl. This system is repre-sented by the following equilibrium equation and equilibrium-constantexpression.

AgCl(s) ⎯→←⎯ Ag+(aq) + Cl−(aq)

The equation represents a heterogeneous reaction, as described onpage 599. Once again, we follow the convention of writing the equilibri-um expression without including the solid species. Therefore, [AgCl]does not appear in the final expression. The resulting equilibriumexpression gives the solubility product constant Ksp. The solubility product constant of a substance is the product of the molar concentra-

tions of its ions in a saturated solution, each raised to the power that is thecoefficient of that ion in the balanced chemical equation.

Ksp = [Ag+][Cl−]

This equation is the solubility equilibrium expression for the reaction. Itexpresses the fact that the solubility product constant, Ksp, of AgCl is theproduct of the molar concentrations of its ions in a saturated solution.

Calcium fluoride is another slightly soluble salt. The equilibrium in asaturated CaF2 solution is described by the following equation.

CaF2(s) ⎯→←⎯ Ca2+(aq) + 2F−(aq)

The solubility product constant has the following form.

Ksp = [Ca2+][F−]2

Notice that Ksp is the product of the molar concentration of Ca2+ ionsand the molar concentration of F− ions squared, as required by the bal-anced chemical equilibrium expression.

The numerical value of Ksp can be determined from solubility data.These data indicate that a maximum of 1.9 × 10−4 g of AgCl can dissolvein 100. g of water at 25°C. One mole of AgCl has a mass of 143.32 g.Thesolubility of AgCl can therefore be expressed in moles per liter ofwater, which is very nearly equal to moles per liter of solution.

× × ×

= 1.3 × 10−5 mol/L

Silver chloride dissociates in solution, contributing equal numbers ofAg+ and Cl− ions. The ion concentrations in the saturated solution aretherefore 1.3 × 10−5 mol/L.

[Ag+] = 1.3 × 10−5

[Cl−] = 1.3 × 10−5

and

Ksp = [Ag+][Cl−]Ksp = (1.3 × 10−5)(1.3 × 10−5)Ksp = (1.3 × 10−5)2

Ksp = 1.7 × 10−10

This result is the solubility product constant of AgCl at 25°C.The solubility of CaF2 is 8.6 × 10−3 g/100 g of water at 25°C. Expressed

in moles per liter, as before, this concentration becomes 1.1 × 10−3

mol/L. CaF2 dissociates in solution to yield twice as many F− ions as

1 mol AgCl

143.32 g AgCl

1000 mL

1 L

1 g H2O

1 mL H2O

1.9 × 10−4 g AgCl

100. g H2O

C H A P T E R 1 8614

Ca2+ ions.The ion concentrations in the saturated solution are 1.1 × 10−3

for the calcium ion and 2(1.1 × 10−3), or 2.2 × 10−3, for the fluoride ion.Note that at equilibrium at 25°C, [Ca2+] equals the solubility of 1.1 ×10−3 mol/L but [F−] equals twice the solubility, or 2.2 × 10−3 mol/L. Thenumber of moles of positive and negative ions per mole of compoundmust always be accounted for when using Ksp and solubilities.

Ksp = [Ca2+][F−]2

Ksp = (1.1 × 10−3)(2.2 × 10−3)2

Ksp = 5.3 × 10−9

Thus, the solubility product constant of CaF2 is 5.3 × 10−9 at 25°C.It is difficult to measure very small concentrations of a solute with

precision. For this reason, solubility data from different sources mayreport different values of Ksp for a substance. Thus, calculations of Kspordinarily should be limited to two significant figures. Representativevalues of Ksp at 25°C for some slightly soluble compounds are listed inTable 3. Assume that all data used in Ksp calculations have been takenat 25°C unless otherwise specified.

At this point, you should note the difference between the solubility ofa given solid and its solubility product constant. Remember that the solubility product constant is an equilibrium constant representing theproduct of the molar concentrations of its ions in a saturated solution. It


Salt Ion product Ksp Salt Ion product Ksp

AgCH3COO [Ag+][CH3COO−] 1.9 × 10−3 CuCl [Cu+][Cl−] 1.2 × 10−6

AgBr [Ag+][Br−] 5.0 × 10−13 CuS [Cu2+][S2−] 6.3 × 10−36

Ag2CO3 [Ag+]2[CO32−] 8.1 × 10−12 FeS [Fe2+][S2−] 6.3 × 10−18

AgCl [Ag+][Cl−] 1.8 × 10−10 Fe(OH)2 [Fe2+][OH−]2 8.0 × 10−16

AgI [Ag+][I−] 8.3 × 10−17 Fe(OH)3 [Fe3+][OH−]3 4 × 10−38

Ag2S [Ag+]2[S2−] 6.3 × 10−50 HgS [Hg2+][S2−] 1.6 × 10−52

Al(OH)3 [Al3+][OH−]3 1.3 × 10−33 MgCO3 [Mg2+][CO32−] 3.5 × 10−8

BaCO3 [Ba2+][CO32−] 5.1 × 10−9 Mg(OH)2 [Mg2+][OH−]2 1.8 × 10−11

BaSO4 [Ba2+][SO42−] 1.1 × 10−10 MnS [Mn2+][S2−] 2.5 × 10−13

CdS [Cd2+][S2−] 8.0 × 10−27 PbCl2 [Pb2+][Cl−]2 1.6 × 10−5

CaCO3 [Ca2+][CO32−] 2.8 × 10−9 PbCrO4 [Pb2+][CrO4

2−] 2.8 × 10−13

CaF2 [Ca2+][F−]2 5.3 × 10−9 PbSO4 [Pb2+][SO42−] 1.6 × 10−8

Ca(OH)2 [Ca2+][OH−]2 5.5 × 10−6 PbS [Pb2+][S2−] 8.0 × 10−28

CaSO4 [Ca2+][SO42−] 9.1 × 10−6 SnS [Sn2+][S2−] 1.0 × 10−25

CoCO3 [Co2+][CO32−] 1.4 × 10−13 SrSO4 [Sr2+][SO4

2−] 3.2 × 10−7

CoS [Co2+][S2−] 4.0 × 10−21 ZnS [Zn2+][S2−] 1.6 × 10−24

TABLE 3 Solubility Product Constants, Ksp, at 25°C

has only one value for a given solid at a given temperature. The solubilityof a solid is an equilibrium position that represents the amount of the solidrequired to form a saturated solution with a specific amount of solvent. Ithas an infinite number of possible values at a given temperature and isdependent on other conditions, such as the presence of a common ion.

C H A P T E R 1 8616

Go to go.hrw.com formore practice problemsthat ask you to calculatesolubility productconstants.

Keyword: HC6EQUX

Given: solubility of CuCl = 1.08 × 10−2 g CuCl/100. g H2OUnknown: Ksp

Start by converting the solubility of CuCl in g/100. g H2O to mol/L. You will need the molarmass of CuCl to get moles CuCl from grams CuCl. Then use the solubility of the [Cu+] and[Cl−] ions in the Ksp expression and solve for Ksp.

× × × = solubility in mol/L

CuCl(s) ⎯→←⎯ Cu+(aq) + Cl−(aq)

Ksp = [Cu+][Cl−]

[Cu+] = [Cl−] = solubility in mol/L

The molar mass of CuCl is 99.0 g/mol.

solubility = × × × =

1.09 × 10−3 mol/L CuCl

[Cu+] = [Cl−] = 1.09 × 10−3 mol/L

Ksp = (1.09 × 10−3)(1.09 × 10−3) = 1.19 × 10−6

The answer contains the proper number of significant figures and is close to the Ksp valuegiven in Table 3.

1 mol CuCl

99.0 g CuCl

1000 mL

1L

1 g H2O

1 mL

1.08 × 10−2 g CuCl

100. g H2O

1 mol CuCl

g CuCl

1000 mL

1 L1 g H2O

1 mL H2O

g CuCl

100. g H2O

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

4 EVALUATE

SAMPLE PROBLEM B

1. Calculate the solubility product constant, Ksp, of lead(II) chloride,PbCl2, which has a solubility of 1.0 g/100. g H2O at 20°C.

2. A 5.0 gram sample of Ag2SO4 will dissolve in 1.0 L of water.Calculate the solubility product constant for this salt.

Answers in Appendix E

Calculate the solubility product constant, Ksp, for copper(I) chloride, CuCl, given that the solubility ofthis compound at 25°C is 1.08 � 10–2 g/100. g H2O.

PRACTICE

Calculating Solubilities

Once known, the solubility product constant can be used to determinethe solubility of a slightly soluble salt. Suppose you wish to know howmany moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C. From Table 3, Ksp for BaCO3 has the numerical value5.1 × 10−9. The equilibrium equation is written as follows.

BaCO3(s) ⎯→←⎯ Ba2+(aq) + CO32−(aq)

Given the value for Ksp, we can write the solubility equilibrium expres-sion as follows.

Ksp = [Ba2+][CO32−] = 5.1 × 10−9

Therefore, BaCO3 dissolves until the product of the molar concentra-tions of Ba2+ ions and CO3

2− ions equals 5.1 × 10−9. The solubility equilibrium equation shows that Ba2+ ions and CO3

2− ions enter thesolution in equal numbers as the salt dissolves.Thus, they have the sameconcentration. Let [Ba2+] = x. Then [CO3

2−] = x also.

[Ba2+][CO32−] = Ksp = 5.1 × 10−9

(x)(x) = x2 = 5.1 × 10−9

x = �5.�1�×� 1�0−�9�

The molar solubility of BaCO3 is 7.1 × 10−5 mol/L.

Thus, the solution concentration is 7.1 × 10−5 M for Ba2+ ions and 7.1 × 10−5 M for CO3

2− ions.


Given: Ksp = 5.0 × 10−13

Unknown: solubility of AgBr

AgBr ⎯→←⎯ Ag+(aq) + Br−(aq)

Ksp = [Ag+][Br−]

[Ag+] = [Br−], so let [Ag+] = x and [Br−] = x

Ksp = [Ag+][Br−]Ksp = x2

x2 = 5.0 × 10−13

x = �5.0 × 1�0−13�Solubility of AgBr = �5.0 × 1�0−13� = 7.1 × 10−7 mol/L

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

SAMPLE PROBLEM C

Calculate the solubility of silver bromide, AgBr, in mol/L, using the Ksp value for this compound listed in Table 3.

Precipitation Calculations

In an earlier example, BaCO3 served as the source of both Ba2+ and CO32−

ions. Because each mole of BaCO3 yields one mole of Ba2+ ions and onemole of CO3

2− ions, the concentrations of the two ions were equal.However, the equilibrium condition does not require that the two ionconcentrations be equal. Equilibrium will still be established so that theion product [Ba2+][CO3

2−] does not exceed the value of Ksp for the system.Similarly, if the ion product [Ca2+][F−]2 is less than the value of Ksp

at a particular temperature, the solution is unsaturated. If the ion productis greater than the value for Ksp, CaF2 precipitates. This precipitationreduces the concentrations of Ca2+ and F− ions until equilibrium isestablished.

Suppose that unequal quantities of BaCl2 and Na2CO3 are dissolvedin water and that the solutions are mixed. If the ion product[Ba2+][CO3

2−] exceeds the Ksp of BaCO3, a precipitate of BaCO3 forms.After precipitation, the ion concentrations are such that [Ba2+][CO3

2−]equals the Ksp.

C H A P T E R 1 8618

Go to go.hrw.com formore practice problemsthat ask you to calculatesolubilities.

Keyword: HC6EQUX

The answer has the proper number of significant figures and is close to an estimated valueof 7.0 × 10−7 calculated as �49 × 10�−14�.

4 EVALUATE

1. Calculate the solubility of cadmium sulfide, CdS, in mol/L, giventhe Ksp value listed in Table 3.

2. Determine the concentration of strontium ions in a saturatedsolution of strontium sulfate, SrSO4, if the Ksp for SrSO4 is 3.2 × 10−7.


(a) (b) (c)

(d) (e) (f)

Ag+

NO3−

Pb2+

NO3− CrO4

2−

Ag+

Pb2+

CrO42−

Ag+

SCN− SCN−

Fe3+FIGURE 13 Nitrate salts of Ag+ (a)and Pb2+ (b) are soluble. When chro-mate ions, CrO4

2−, combine with Ag+

(c) or Pb2+ (d), a slightly soluble saltforms. Thiocyanate ions, SCN−, canform a slightly soluble salt with Ag+

(e) or a soluble salt with Fe3+ (f).

Substances differ greatly in their tendencies to form precipitateswhen mixed in moderate concentrations. The photos in Figure 13 showthe behavior of some anions in the presence of certain cations. Note thatsome of the combinations have produced precipitates and some havenot. The solubility product can be used to predict whether a precipitateforms when two solutions are mixed.


Given: concentration of BaCl2 = 0.010 Mvolume of BaCl2 = 20.0 mLconcentration of Na2SO4 = 0.0050 Mvolume of Na2SO4 = 20.0 mL

Unknown: whether a precipitate forms

The two possible new pairings of ions are NaCl and BaSO4. Of these, only BaSO4 is aslightly soluble salt. It will precipitate if the ion product [Ba2+][SO4

2−] in the mixed solutionexceeds Ksp for BaSO4. From the list of solubility products in Table 3, the Ksp is found to be 1.1 × 10−10. The solubility equilibrium equation follows.

BaSO4(s) ⎯→←⎯ Ba2+(aq) + SO42−(aq)

The solubility equilibrium expression is written as follows.

Ksp = [Ba2+][SO42−] = 1.1 × 10−10

First [Ba2+] and [SO42−] in the above solution must be found. Then the ion product is

calculated and compared with the Ksp.

Calculate the mole quantities of Ba2+ and SO42− ions.

0.020 L × = 0.000 20 mol Ba2+

0.020 L × = 0.000 10 mol SO42−

Calculate the total volume of solution containing Ba2+ and SO42− ions.

0.020 L + 0.020 L = 0.040 L

Calculate the Ba2+ and SO42− ion concentrations in the combined solution.

= 5.0 × 10−3 mol/L Ba2+

= 2.5 × 10−3 mol/L SO42−0.000 10 mol SO4

2−

0.040 L

0.000 20 mol Ba2+

0.040 L

0.0050 mol SO42−

1 L

0.010 mol Ba2+

1 L

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

SAMPLE PROBLEM D

Will a precipitate form if 20.0 mL of 0.010 M BaCl2 is mixed with 20.0 mL of 0.0050 M Na2SO4?

Limitations on the Use of Ksp

The solubility product principle can be very useful when applied to solu-tions of slightly soluble substances. It cannot be applied to solutions of sol-uble substances. This is because the positive and negative ions attract eachother, and this attraction becomes appreciable when the ions are closetogether. Sometimes it is necessary to consider two equilibria simultane-ously. For example, if either ion hydrolyzes, the salt will be more solublethan predicted when only the solubility product constant is used.The solu-bility product is also sensitive to changes in solution temperature to theextent that the solubility of the dissolved substance is affected by suchchanges.All of these factors limit the conditions under which the solubilityproduct principle can be applied.

C H A P T E R 1 8620

Go to go.hrw.com formore practice problemsthat ask you to performprecipitationcalculations.

Keyword: HC6EQUX

Calculate the ion product.

[Ba2+][SO42−] = (5.0 × 10−3)(2.5 × 10−3)

= 1.2 × 10−5

The ion product, 1.2 × 10−5, is greater than the value of Ksp, 1.1 × 10−10, so precipitationoccurs.

The answer contains the appropriate number of significant figures and is close to an esti-mated value of 1 × 10−5, calculated as (5 × 10−3)(2 × 10−3); because 10−5 > 10−10, precipitationshould occur.

4 EVALUATE

1. Does a precipitate form when 100. mL of 0.0025 M AgNO3and 150. mL of 0.0020 M NaBr solutions are mixed?

2. Does a precipitate form when 20. mL of 0.038 M Pb(NO3)2and 30. mL of 0.018 M KCl solutions are mixed?


1. What is a solubility product constant? How aresuch constants determined?

2. How are solubility product constants used to calculate solubilities?

3. What is an ion product?

4. How are calculations to predict possible precipita-tion carried out?

5. What is the value of Ksp for Ag2SO4 if 5.40 g is soluble in 1.00 L of water?

6. Determine whether a precipitate will form if20.0 mL of 1.00 × 10−7 M AgNO3 is mixed with 20.0 mL of 2.00 × 10−9 M NaCl at 25°C.

Critical Thinking

7. ANALYZING DATA A solution is 0.20 M in each ofthe following: Ca(NO3)2, Cr(NO3)3, and La(NO3)3.Solid NaF is added to the solution until the [F–] of the solution is 1.0 × 10–4 M. Given the valuesof Ksp below, describe what will happen.CaF2 = 3.9 × 10–11; CrF3 = 6.6 × 10–11; andLaF3 = 4.0 × 10–17

SECTION REVIEW



• According to Le Châtelier’s principle, when a stress (a changein concentration, pressure, or temperature) is applied to a sys-tem at equilibrium, the equilibrium is shifted in the directionthat relieves the stress.

• The common-ion effect is recognized when a solution contain-ing ions such as those of a reactant or a product in an equilibri-um system is added to the system. Le Châtelier’s principleexplains the response of the system to the stress.

• A reaction system in which the forward and reverse reactionsoccur simultaneously and at the same rate is said to be at equi-librium. Both reactions continue, but there is no net change inthe composition of the system.

• At equilibrium, the ratio of the product of the molar concen-trations of substances formed to the product of the molar con-centrations of reactants, each raised to the appropriate power,has a definite numerical value, K, which is the equilibrium con-stant at a given temperature.

reversible reactionchemical equilibriumequilibrium constantchemical equilibrium expression

common-ion effect

Vocabulary

Vocabulary

The Nature of Chemical Equilibrium

Shifting Equilibrium

• Ions of salts that are slightly soluble form saturated aqueoussolutions at low concentrations. The solubility equilibriumexpression for such salts yields a constant—the solubility productconstant, Ksp.

solubility product constantVocabulary

Solubility Equilibrium

• The equilibrium expression for the ionization constant of theweak acid HA follows.

Ka =

• Salts formed from strong bases and weak acids produce aque-ous solutions that are basic because of anion hydrolysis.

• Salts formed from strong acids and weak bases produce aque-ous solutions that are acidic because of cation hydrolysis.

• Salts formed from strong acids and strong bases do nothydrolyze in water, and their solutions are neutral.

• Salts formed from weak acids and weak bases may produceneutral, acidic, or basic solutions, depending on the relativeamounts of cation and anion hydrolysis.

[H3O+][A−]

[HA]

acid ionization constantbuffered solutionhydrolysis

Vocabulary

Equilibria of Acids, Bases, and Salts

C H A P T E R R E V I E WFor more practice, go to the Problem Bank in Appendix D.

C H A P T E R 1 8622

The Nature of ChemicalEquilibriumSECTION 1 REVIEW

1. Describe and explain how the concentrations ofA, B, C, and D change from the time when Aand B are first combined to the point at whichequilibrium is established for the reactionA + B ⎯→←⎯ C + D.

2. a. Write the general expression for an equilibri-um constant based on the equation nA +mB + . . . ⎯→←⎯ xC + yD + . . .

b. What information is provided by the value ofK for a given equilibrium system at a speci-fied temperature?

3. In general, which reaction is favored (forwardor reverse) if the value of K at a specified tem-perature isa. very small?b. very large?

PRACTICE PROBLEMS

4. Determine the value of the equilibrium con-stant for each reaction given, assuming that theequilibrium concentrations are found to bethose specified. (Concentrations are in mol/L.)(Hint: See Sample Problem A.)a. A + B ⎯→←⎯ C; [A] = 2.0; [B] = 3.0; [C] = 4.0b. D + 2E ⎯→←⎯ F + 3G; [D] = 1.5; [E] = 2.0;

[F] = 1.8; [G] = 1.2c. N2(g) + 3H2(g) ⎯→←⎯ 2NH3(g); [N2] = 0.45;

[H2] = 0.14; [NH3] = 0.625. An equilibrium mixture at a specific tempera-

ture is found to consist of 1.2 × 10−3 mol/L HCl,3.8 × 10−4 mol/L O2, 5.8 × 10−2 mol/L H2O, and5.8 × 10−2 mol/L Cl2 according to the following:

4HCl(g) + O2(g) ⎯→←⎯ 2H2O(g) + 2Cl2(g).Determine the value of the equilibrium con-stant for this system.

6. At 450°C, the value of the equilibrium constantfor the following system is 6.59 × 10−3. If[NH3] = 1.23 × 10−4 M and [H2] = 2.75 × 10−2 M

at equilibrium, determine the concentration ofN2 at that point.

N2(g) + 3H2(g) ⎯→←⎯ 2NH3(g)7. The value of the equilibrium constant for the

reaction below is 40.0 at a specified temperature.What would be the value of that constant for thereverse reaction under the same conditions?

H2(g) + I2(g) ⎯→←⎯ 2HI(g)

Shifting EquilibriumSECTION 2 REVIEW

8. Predict whether each of the following pressurechanges would favor the forward or reversereaction.

2NO(g) + O2(g) ⎯→←⎯ 2NO2(g)a. increased pressureb. decreased pressure

9. In heterogeneous reaction systems, what typesof substances do not appear in the equilibriumconstant expression? Why?

10. Explain the effect of a catalyst on an equilibri-um system.

11. Predict the effect of each of the following onthe indicated equilibrium system in terms of thedirection of equilibrium shift (forward, reverse,or neither).

H2(g) + Cl2(g) ⎯→←⎯ 2HCl(g) + 184 kJa. addition of Cl2b. removal of HClc. increased pressured. decreased temperaturee. removal of H2f. decreased pressureg. addition of a catalysth. increased temperaturei. decreased system volume

12. How would the changes in (a) through (i) ofitem 11 affect the new equilibrium concentra-tion of HCl and the value of K at the new equilibrium?

13. Explain why changes in the concentrations of thereactants and products at equilibrium have noimpact on the value of the equilibrium constant.

CHAPTER REVIEW


14. What relative pressure (high or low) wouldresult in the production of the maximum levelof CO2 according to the following equation?Why?

2CO(g) + O2(g) ⎯→←⎯ 2CO2(g)15. What relative conditions (reactant concentra-

tions, pressure, and temperature) would favor ahigh equilibrium concentration of the under-lined substance in each of the following equilib-rium systems?a. 2CO(g) + O2(g) ⎯→←⎯ + 167 kJ

b. Cu2+(aq) + 4NH3(aq) ⎯→←⎯+ 42 kJ

c. 2HI(g) + 12.6 kJ ⎯→←⎯ H2(g) +

d. 4HCl(g) + O2(g) ⎯→←⎯2H2O(g) + + 113 kJ

e. PCl5(g) + 88 kJ ⎯→←⎯ PCl3(g) +

16. The reaction between hemoglobin, Hb, and oxy-gen, O2, in red blood cells is responsible fortransporting O2 to body tissues. This process canbe represented by the following equilibriumreaction:

Hb(aq) + O2(g) ⎯→←⎯ HbO2(aq)What will happen to the concentration of oxy-genated hemoglobin, HbO2, at high altitude,where the pressure of oxygen is 0.1 atm insteadof 0.2 atm, as it is at sea level?

17. What two factors determine the extent to whichreacting ions are removed from solution?

18. Identify the three conditions under which ionicreactions can run to completion, and write anequation for each.

Equilibria of Acids, Bases, and SaltsSECTION 3 REVIEW

19. a. Write the ion product constant expressionfor water.

b. What is the value of this constant at 25°C?20. List and distinguish between the four general

categories of salts, based on their hydrolysisproperties, and give an example of each.

Cl2(g)

2Cl2(g)

I2(g)

Cu(NH3)42+(aq)

2CO2(g)

21. Explain why the pH of a solution containingboth acetic acid and sodium acetate is higherthan that of a solution containing the same con-centration of acetic acid alone.

22. The ionization constant, Ka, for acetic acid is1.8 × 10−5 at 25°C. Explain the significance ofthis value.

23. a. From the development of Ka described inSection 3, show how you would express an ion-ization constant, Kb, for the weak base NH3.

b. In this case, Kb = 1.8 × 10−5. What is the signifi-cance of this numerical value to equilibrium?

Solubility EquilibriumSECTION 4 REVIEW

24. Explain why a saturated solution is not neces-sarily a concentrated solution.

25. What rule of thumb is used to distinguishbetween soluble and slightly soluble substances?

26. What is the relative ion concentration of anionic substance typically involved in solubilityequilibrium systems?

27. What is the relationship between Ksp and theproduct of the ion concentrations in terms ofdetermining whether a solution of those ions is saturated?

PRACTICE PROBLEMS

28. The ionic substance EJ dissociates to form E2+ and J2− ions. The solubility of EJ is 8.45 ×10−6 mol/L. What is the value of the solubilityproduct constant? (Hint: See Sample Problem B.)

29. Calculate the solubility product constant Ksp foreach of the following, based on the solubilityinformation provided:a. BaSO4 = 2.4 × 10−4 g/100. g H2O at 20°Cb. Ca(OH)2 = 0.173 g/100. g H2O at 20°C

30. Calculate the molar solubility of a substanceMN that ionizes to form M2+ and N2− ions,given that Ksp = 8.1 × 10−6. (Hint: See SampleProblem C.)

CHAPTER REVIEW

31. Use the Ksp values given in Table 3 to evaluatethe solubility of each of the following in molesper liter.a. AgBrb. CoS

32. Complete each of the following relative to the reaction that occurs when 25.0 mL of0.0500 M Pb(NO3)2 is combined with 25.0 mLof 0.0400 M Na2SO4 if equilibrium is reached at 25°C.a. Write the solubility equilibrium equation at

25°C.b. Write the solubility equilibrium expression

for the net reaction.33. The ionic substance T3U2 ionizes to form T2+

and U3− ions. The solubility of T3U2 is 3.8 ×10−10 mol/L. What is the value of the solubilityproduct constant?

34. A solution of AgI contains 2.7 × 10−10 mol/L Ag+.What is the maximum I− concentration that canexist in this solution?

35. Calculate whether a precipitate will form if 0.35 L of 0.0044 M Ca(NO3)2 and 0.17 L of0.000 39 M NaOH are mixed at 25°C. (See Table 3for Ksp values.) (Hint: See Sample Problem D.)

36. Determine whether a precipitate will form if1.70 g of solid AgNO3 and 14.5 g of solid NaClare dissolved in 200. mL of water to form asolution at 25°C.

37. If 2.50 × 10−2 g of solid Fe(NO3)3 is added to100. mL of a 1.0 × 10−4 M NaOH solution, will aprecipitate form?

38. Calcium carbonate is only slightly soluble inwater.a. Write the equilibrium equation for calcium

carbonate in solution.b. Write the solubility product constant expres-

sion, Ksp, for the equilibrium in a saturatedsolution of CaCO3.

39. Calculate the concentration of Hg2+ ions in asaturated solution of HgS(s). How many Hg2+

ions are in 1000 L of the solution?

MIXED REVIEW

40. Calculate the equilibrium constant, K, for thefollowing reaction at 900°C.

H2(g) + CO2(g) ⎯→←⎯ H2O(g) + CO(g)The components were analyzed, and it was foundthat [H2] = 0.061 mol/L, [CO2] = 0.16 mol/L,[H2O] = 0.11 mol/L, and [CO] = 0.14 mol/L.

41. A solution in equilibrium with solid bariumphosphate is found to have a barium ion concen-tration of 5.0 × 10−4 M and a Ksp of 3.4 × 10−23.Calculate the concentration of phosphate ion.

42. At 25°C, the value of K is 1.7 × 10−13 for the fol-lowing reaction.

2N2O(g) + O2(g) ⎯→←⎯ 4NO(g)It is determined that [N2O] = 0.0035 mol/L and[O2] = 0.0027 mol/L. Using this information, whatis the concentration of NO(g) at equilibrium?

43. Tooth enamel is composed of the mineralhydroxyapatite, Ca5(PO4)3OH, which has a Kspof 6.8 × 10−37. The molar solubility of hydroxy-apatite is 2.7 × 10−5 mol/L. When hydroxyap-atite is reacted with fluoride, the OH− isreplaced with the F− ion on the mineral, form-ing fluorapatite, Ca5(PO4)3F. (The latter is hard-er and less susceptible to cavities.) The Ksp offluorapatite is 1 × 10−60. Calculate the molar sol-ubility of fluorapatite in water. Given your cal-culations, can you support the fluoridation ofdrinking water?

44. Determine if a precipitate will form when0.96 g Na2CO3 is combined with 0.20 g BaBr2in a 10. L solution (Ksp = 2.8 × 10−9).

45. For the formation of ammonia, the equilibriumconstant is calculated to be 5.2 × 10−5 at 25°C.After analysis, it is determined that [N2] = 2.00 Mand [H2] = 0.80 M. How many grams of ammoniaare in the 10. L reaction vessel at equilibrium?Use the following equilibrium equation.

N2(g) + 3H2(g) ⎯→←⎯ 2NH3(g)

46. Relating Ideas Let s equal the solubility, inmol/L, of AB2. In terms of s, what is the molarconcentration of A? of B? What is the Ksp ofAB2?

CRITICAL THINKING

C H A P T E R 1 8624

CHAPTER REVIEW


47. Predicting Outcomes When gasoline burns inan automobile engine, nitric oxide is formedfrom oxygen and nitrogen. Nitric oxide is amajor air pollutant. High temperatures such asthose found in a combustion engine are neededfor the following reaction:

N2(g) + O2(g) ⎯→←⎯ 2NO(g)K for the reaction is 0.01 at 2000°C. If 4.0 molof N2, 0.1 mol of O2, and 0.08 mol of NO areplaced in a 1.0 L vessel at 2000°C, predict whichreaction will be favored.

48. An equilibrium system helps maintain the pHof the blood. Review the material on the carbondioxide–bicarbonate ion equilibrium system inGroup 14 of the Elements Handbook, andanswer the following.a. Write the equation for the equilibrium

system that responds to changes in H3O+

concentration.b. Use Le Châtelier’s principle to explain how

hyperventilation affects this system.c. How does this system maintain pH when

acid is added?49. The reactions used to confirm the presence of

transition metal ions often involve the formationof precipitates. Review the analytical tests for thetransition metals in the Elements Handbook. Usethat information and Table 3 to determine theminimum concentration of Zn2+ needed to pro-duce a precipitate that confirms the presence ofZn. Assume enough sulfide ion reagent is addedto the unknown solution in the test tube to pro-duce a sulfide ion concentration of 1.4 × 10−20 M.

50. Find photos of several examples of stalagmitesand stalactites in various caves. Investigate theequilibrium processes involved in the formationof stalagmites and stalactites.

RESEARCH & WRITING

USING THE HANDBOOK

51. Carry out library research on the use of cata-lysts in industrial processes. Explain what typesof catalysts are used for specific processes, suchas the Haber process.

52. Research nitrogen narcosis in the library. Whatcauses nitrogen narcosis, and how does it relateto Le Châtelier’s principle?


Graphing Calculator ChemicalEquilibrium

Go to go.hrw.com for a graphing calculatorexercise that asks you to calculate the per-cent ionization for an acid equilibrium.

Keyword: HC6EQUX

Problem Solving S-

• Always use a balanced chemical equation to write an equilibrium-constant equation.

• To write an equation, place the product concentrations in the numerator and thereactant concentrations in the denominator. Raise each substance’s concentrationto the power equal to the substance’s coefficient in the balanced chemical equation.

• The concentration of any solid or pure liquid that takes part in the reaction is leftout because these concentrations never change.

SAMPLE

PRACTICE PROBLEMS

For a given temperature, you can write a mathematical equation that describes the equi-librium of a reaction in terms of concentration. The equation defines an equilibrium con-stant, K, as a function of the concentrations of products and reactants at equilibrium.

Consider an equilibrium process in which reactants A and B form products C and D.nA + mB ⎯→←⎯ xC + yD

The terms n, m, x, and y are the coefficients of the balanced equation.

K =[C]x[D]y⎯[A]n[B]m

Write an equation for the equilibrium constant of the reaction in which nitrogen monoxidechanges to dinitrogen monoxide and nitrogen dioxide.

To write an equation for an equilibrium constant, you must start with a balanced chemicalequation for the equilibrium reaction. By writing the formulas of the compounds mentioned inthe description, you get the unbalanced equilibrium equation NO(g) ⎯→←⎯ N2O(g) + NO2(g).

Balancing the equation requires a coefficient of 3 in front of NO, giving 3NO(g) ⎯→←⎯N2O(g) + NO2(g). Next, write an equilibrium equation. Remember, each concentration in theequilibrium equation is raised to a power equal to its coefficient in the balanced chemical equa-tion.The product concentrations, [N2O] and [NO2], are placed in the numerator.The coefficientof each of the products is 1, so the exponent of each concentration is 1. There is only one reac-tant, so its concentration, [NO], is written in the denominator. Its coefficient is 3 in the balancedchemical equation, so the concentration of NO is raised to the third power.The exponents witha value of 1 do not have to be written. The resulting equation is

K = =[N2O][NO2]⎯⎯

[NO]3

[N2O]1[NO2]1⎯⎯

[NO]3

C H A P T E R 1 8626

Math Tutor DETERMINING EQUILIBRIUM CONSTANTS

1. Write equations for the equilibrium constant ofeach of the following hypothetical reactions:a. A(aq) + 2B(aq) ⎯→←⎯ AB2(aq)b. 2DE2(g) ⎯→←⎯ D2(g) + 2E2(g)

2. Use the equilibrium concentrations below to cal-culate the equilibrium constant for the followingdecomposition reaction:

2BrF5(g) ⎯→←⎯ Br2(g) + 5F2(g)[BrF5] = 0.000137 mol/L, [Br2] = 0.00050 mol/L,and [F2] = 0.0025 mol/L



MULTIPLE CHOICE

1.A chemical reaction is in equilibrium whenA. forward and reverse reactions have ceased.B. the equilibrium constant equals 1.C. forward and reverse reaction rates are equal.D. No reactants remain.

2.Which change can cause the value of the equi-librium constant to change?A. temperatureB. concentration of a reactantC. concentration of a productD. None of the above

3.Consider the following reaction:2C(s) + O2(g) ⎯→←⎯ 2CO(g)

The equilibrium constant expression for thisreaction is

A. . C. .

B. . D. .

4.The solubility product of cadmium carbonate,CdCO3, is 1.0 × 10−12. In a saturated solution ofthis salt, the concentration of Cd2+(aq) ions isA. 5.0 × 10−13 mol/L. C. 1.0 × 10−6 mol/L.B. 1.0 × 10−12 mol/L. D. 5.0 × 10−7 mol/L.

5.Consider the following equation for an equilib-rium system:2PbS(s) + 3O2(g) + C(s) ⎯→←⎯

2Pb(s) + CO2(g) + 2SO2(g)Which concentration(s) would be included inthe denominator of the equilibrium constantexpression?A. Pb(s), CO2(g), and SO2(g)B. PbS(s), O2(g), and C(s)C. O2(g), Pb(s), CO2(g), and SO2(g)D. O2(g)

6.If an exothermic reaction has reached equilib-rium, then increasing the temperature willA. favor the forward reaction.B. favor the reverse reaction.C. favor both the forward and reverse reactions.D. have no effect on the equilibrium.

[CO]⎯[O2]2

[CO]2⎯[O2][C]2

2[CO]⎯[O2][2C]

[CO]2⎯[O2]

7.Le Châtelier’s principle states thatA. at equilibrium, the forward and reverse reac-

tion rates are equal.B. stresses include changes in concentrations,

pressure, and temperature.C. to relieve stress, solids and solvents are omit-

ted from equilibrium constant expressions.D. chemical equilibria respond to reduce

applied stress.

SHORT ANSWER

8.Describe the conditions that would allow you toconclusively determine that a solution is satu-rated. You can use only visual observation andcannot add anything to the solution.

9.The graph below shows the neutralization curvefor 100 mL of 0.100 M acid with 0.100 M base.Which letter represents the equivalence point?What type of acid and base produced thiscurve?


Keeping a positive attitude dur-ing any test will help you focus on the test and likelyimprove your score.

mL base added

pH

13

11

9

7

5

3

10

0 10 30 50 70 90 110 130 150

a

c

b

d

EXTENDED RESPONSE

10.Explain how the same buffer can resist a changein pH when either an acid or a base is added.Give an example.

Measuring Ka for Acetic Acid

CHAPTER LAB MICROL A B

C H A P T E R 1 8628

BACKGROUNDThe acid dissociation constant, Ka, is a measure of the strength of an acid. Strong acids are complete-ly ionized in water. Because weak acids are onlypartly ionized, they have a characteristic Ka value.Properties that depend on the ability of a substanceto ionize, such as conductivity and colligative proper-ties, can be used to measure Ka. In this experiment,you will compare the conductivity of a 1.0 M solu-tion of acetic acid, CH3COOH, a weak acid, with theconductivities of solutions of varying concentrationsof hydrochloric acid, HCl, a strong acid. From thecomparisons you make, you will be able to estimatethe concentration of hydronium ions in the aceticacid solution and calculate its Ka.

SAFETY

For review of safety, please see Safety in theChemistry Laboratory in the front of your book.

PREPARATION1. Create a table with two columns for recording

your observations. Head the first column “HClconcentration.” A wide second column can beheaded “Observations and comparisons.”

PROCEDURE1. Obtain samples of 1.0 M HCl solution and 1.0 M

CH3COOH solution.

2. Place 20 drops of HCl in one well of a 24-wellplate. Place 20 drops of CH3COOH in an adja-cent well. Label the location of each sample.

3. Test the HCl and CH3COOH with the conduc-tivity tester. Note the relative intensity of the

OBJECTIVES

• Compare the conductivities of solutions of known and unknown hydronium ionconcentrations.

• Relate conductivity to the concentration of ions in solution.

• Explain the validity of the procedure on the basis of the definitions of strong and weak acids.

• Compute the numerical value of Ka foracetic acid.

MATERIALS

• 1.0 M acetic acid, CH3COOH

• 1.0 M hydrochloric acid, HCl

• 24-well plate

• distilled or deionized water

• LED conductivity testers

• paper towels

• thin-stemmed pipets

Teacher-made LED conductivity tester

Film canistercap

Film canister

LED

20 cmwire

10 cmwire

CLEANUP AND DISPOSAL9. Clean your lab station. Clean all equip-

ment, and return it to its proper place.Dispose of chemicals and solutions incontainers designated by your teacher. Do notpour any chemicals down the drain or throw any-thing in the trash unless your teacher directs youto do so. Wash your hands thoroughly after allwork is finished and before you leave the lab.

ANALYSIS AND INTERPRETATION1. Resolving Discrepancies: How did the conduc-

tivity of the 1.0 M HCl solution compare withthat of the 1.0 M CH3COOH solution? Why doyou think this was so?

2. Organizing Data: What is the H3O+ concentra-tion of the HCl solution that most closelymatched the conductivity of the acetic acid?

3. Drawing Conclusions: What was the H3O+

concentration of the 1.0 M CH3COOH solution? Why?

CONCLUSIONS1. Applying Models: The acid ionization expression

for CH3COOH is the following:

Ka =

Use your answer to Analysis and Interpretationitem 3 to calculate Ka for the acetic acid solution.

2. Applying Models: Explain how it is possiblefor solutions of HCl and CH3COOH to showthe same conductivity but have different concentrations.

EXTENSIONS

1. Evaluating Methods: Compare the Ka value that you calculated with the value found on page 606 of your text. Calculate the percenterror for this experiment.

2. Predicting Outcomes: Lactic acid (HOOCCHOHCH3) has a Ka of 1.4 × 10−4. Predictwhether a solution of lactic acid would cause theconductivity tester to glow brighter or dimmer thana solution of acetic acid with the same concentra-tion. How noticeable would the difference be?

[H3O+][CH3COO−]

[CH3COOH]

tester light for each solution. After testing, rinsethe tester probes with distilled water. Removeany excess moisture with a paper towel.

4. Place 18 drops of distilled water in each of sixwells in your 24-well plate. Add 2 drops of 1.0 M HCl to the first well to make a total of 20 drops of solution. Mix the contents of thiswell thoroughly by picking the contents up in apipet and returning them to the well.

5. Repeat this procedure by taking 2 drops of theprevious dilution and placing it in the next wellcontaining 18 drops of water. Return any unusedsolution in the pipet to the well from which itwas taken. Mix the new solution with a newpipet. (You now have 1.0 M HCl in the well fromProcedure step 2, 0.10 M HCl in the first dilutionwell, and 0.010 M HCl in the second dilution.)

6. Continue diluting in this manner until you havesix successive dilutions. The [H3O+] should nowrange from 1.0 M to 1.0 × 10−6 M. Write the con-centrations in the first column of your data table.

7. Using the conductivity tester, test the cells con-taining HCl in order from most concentrated toleast concentrated. Note the brightness of thetester bulb, and compare it with the brightness ofthe bulb when it was placed in the acetic acidsolution. (Retest the acetic acid well any time forcomparison.) After each test, rinse the testerprobes with distilled water, and use a paper towelto remove any excess moisture. When the bright-ness produced by one of the HCl solutions isabout the same as that produced by the aceticacid, you can infer that the two solutions haveabout the same hydronium ion concentration andthat the pH of the HCl solution is equal to thepH of the acetic acid. If the glow from the bulb istoo faint to see, turn off the lights or build a lightshield around your conductivity tester bulb.

8. Record the results of your observations by notingwhich HCl concentration causes the intensity ofthe bulb to most closely match that of the bulbwhen it is in acetic acid. (Hint: If the conductivityof no single HCl concentration matches that ofthe acetic acid, then estimate the value betweenthe two concentrations that match the best.)


Oxidation-ReductionReactions

Copper oxidizes in air to form the green patina you see on this building.

C H A P T E R 1 9

Columbus Tower, San FranciscoColumbus Tower, San Francisco

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 631

SECTION 1

OBJECTIVES

Assign oxidation numbers toreactant and product species.

Define oxidation andreduction.

Explain what an oxidation-reduction reaction (redoxreaction) is.

Oxidation andReduction

O xidation-reduction reactions involve a transfer of electrons. Oxida-tion involves the loss of electrons, whereas reduction involves the gainof electrons. Reduction and oxidation half-reactions must occur simul-taneously. These processes can be identified through the understandingand use of oxidation numbers (oxidation states).

Oxidation States

Oxidation states were defined in Chapter 7. The oxidation numberassigned to an element in a molecule is based on the distribution of elec-trons in that molecule.The rules by which oxidation numbers are assignedwere given in Chapter 7. These rules are summarized in Table 1.

Rule Example

1. The oxidation number of any pure element is 0. The oxidation number of Na(s) is 0.

2. The oxidation number of a monatomic ion equals the The oxidation number ofcharge on the ion. Cl− is −1.

3. The more electronegative element in a binary compound is assigned The oxidation numberthe number equal to the charge it would have if it were an ion. of O in NO is −2.

4. The oxidation number of fluorine in a compound is always −1. The oxidation number ofF in LiF is −1.

5. Oxygen has an oxidation number of −2 unless it is combined with F, The oxidation numberin which it is +1 or +2, or it is in a peroxide, in which it is −1. of O in NO2 is −2.

6. Hydrogen’s oxidation state in most of its compounds is +1 The oxidation numberunless it is combined with a metal, in which case it is −1. of H in LiH is −1.

7. In compounds, Group 1 and 2 elements and aluminum have The oxidation numberoxidation numbers of +1, +2, and +3, respectively. of Ca in CaCO3 is +2.

8. The sum of the oxidation numbers of all atoms The oxidation number ofin a neutral compound is 0. C in CaCO3 is +4.

9. The sum of the oxidation numbers of all atoms in a The oxidation number ofpolyatomic ion equals the charge of the ion. P in H2PO4

− is +5.

TABLE 1 Rules for Assigning Oxidation Numbers

Chromium provides a very visual example of oxidation numbers.Different oxidation states of chromium have dramatically different col-ors, as can be seen in Figure 1. Solutions with the same oxidation stateshow less dramatic differences. The chromium(II) chloride solution isblue, chromium(III) chloride solution is green, potassium chromatesolution is yellow, and potassium dichromate solution is orange.

Oxidation

Processes in which the atoms or ions of an element experience anincrease in oxidation state are oxidation processes. The combustion ofmetallic sodium in an atmosphere of chlorine gas is shown in Figure 2.The sodium ions and chloride ions produced during this stronglyexothermic reaction form a cubic crystal lattice in which sodium cationsare ionically bonded to chloride anions. The chemical equation for thisreaction is written as follows.

2Na(s) + Cl2(g) ⎯→ 2NaCl(s)

The formation of sodium ions illustrates an oxidation process becauseeach sodium atom loses an electron to become a sodium ion. The oxi-dation state is represented by placing an oxidation number above thesymbol of the atom and the ion.

The oxidation state of sodium has changed from 0, its elementalstate, to the +1 state of the ion (Rules 1 and 7, Table 1). A species whoseoxidation number increases is oxidized. The sodium atom is oxidized toa sodium ion.

0 +1Na ⎯→ Na+ + e−

C H A P T E R 1 9632

+2 +3 +6 +6

Oxidation state

FIGURE 1 The color of solutionscontaining chromium compoundschanges with the oxidation state ofchromium.

FIGURE 2 Sodium and chlorinereact violently to form NaCl. Thesynthesis of NaCl from its elementsillustrates the oxidation-reductionprocess.

Reduction

Processes in which the oxidation state of an element decreases are reduc-tion processes. Consider the behavior of chlorine in its reaction withsodium. Each chlorine atom accepts an electron and becomes a chlorideion. The oxidation state of chlorine decreases from 0 to −1 for the chlo-ride ion (Rules 1 and 2, Table 1).

A species that undergoes a decrease in oxidation state is reduced. Thechlorine atom is reduced to the chloride ion.

Oxidation and Reduction as a Process

Electrons are released in oxidation and acquired in reduction. There-fore, for oxidation to occur during a chemical reaction, reduction mustalso occur. Furthermore, the number of electrons produced in oxidationmust equal the number of electrons acquired in reduction. This makessense when you recall that electrons are negatively charged and that forcharge to be conserved, the number of electrons lost must equal thenumber of electrons gained. You learned in Chapter 8 that mass is con-served in any chemical reaction. Therefore, the masses of the elementsthat undergo oxidation and reduction and the electrons that areexchanged are conserved.

A transfer of electrons causes changes in the oxidation states of one ormore elements. Any chemical process in which elements undergo changesin oxidation number is an oxidation-reduction reaction. This name isoften shortened to redox reaction. An example of a redox reaction can beseen in Figure 3, in which copper is being oxidized and NO3

− from nitricacid is being reduced. The part of the reaction involving oxidation orreduction alone can be written as a half-reaction. The overall equation fora redox reaction is the sum of two half-reactions. Because the number ofelectrons involved is the same for oxidation and reduction, they canceleach other out and do not appear in the overall chemical equation.Equations for the reaction between nitric acid and copper illustrate therelationship between half-reactions and the overall redox reaction.

(oxidation half-reaction)

(reduction half-reaction)

(redox reaction)

Notice that electrons lost in oxidation appear on the product side ofthe oxidation half-reaction. Electrons are gained in reduction and

0 +5 +2 +43Cu + 2NO3

− + 4H+ ⎯→ Cu2+ + 2NO2 + 2H2O

+5 −2 +1 +4 −2 +1 −22NO3

− + 2e− + 4H+ ⎯→ 2NO2 + 2H2O

0 +23Cu ⎯→ Cu2+ + 2e−

0 −1Cl2 + 2e− ⎯→ 2Cl−


FIGURE 3 Copper is oxidized andnitrogen dioxide is produced whenthis penny is placed in a concentrat-ed nitric acid solution.

www.scilinks.orgTopic: Redox ReactionsCode: HC61281

appear as reactants in the reduction half-reaction. When metallic copperreacts in nitric acid, one copper atom is oxidized to Cu2+ as two nitro-gen atoms are reduced from a +5 oxidation state to a +4 oxidation state.Atoms are conserved. This is illustrated by the balanced chemical equa-tion for the reaction between copper and nitric acid.

If none of the atoms in a reaction change oxidation state, the reactionis not a redox reaction. For example, sulfur dioxide gas, SO2, dissolves inwater to form an acidic solution containing a low concentration of sul-furous acid, H2SO3.

The oxidation states of all elemental species remain unchanged in thiscomposition reaction. Therefore, it is not a redox reaction.

When a solution of sodium chloride is added to a solution of silvernitrate, an ion-exchange reaction occurs and white silver chlorideprecipitates.

+ + + ⎯→ + +

The oxidation state of each monatomic ion remains unchanged. Again,this reaction is not an oxidation-reduction reaction.

Redox Reactions and Covalent BondsBoth the synthesis of NaCl from its elements and the reaction betweencopper and nitric acid involve ionic bonding. Substances with covalentbonds also undergo redox reactions. An oxidation number, unlike anionic charge, has no physical meaning. That is, the oxidation numberassigned to a particular atom is based on its electronegativity relative tothe other atoms to which it is bonded in a given molecule; it is not basedon any real charge on the atom. For example, an ionic charge of 1−results from the complete gain of one electron by an atom or other neu-tral species, whereas an oxidation state of −1 means an increased attrac-tion for a bonding electron. A change in oxidation number does notrequire a change in actual charge.

When hydrogen burns in chlorine, a covalent bond forms from thesharing of two electrons. The two bonding electrons in the hydrogenchloride molecule are not shared equally. Rather, the pair of electronsis more strongly attracted to the chlorine atom because of its higherelectronegativity.

+ ⎯→

As specified by Rule 3 in Table 1, chlorine in HCl is assigned an oxida-tion number of −1. Thus, the oxidation number for the chlorine atomschanges from 0, its oxidation number in the elemental state, to −1; chlo-rine atoms are reduced. As specified by Rule 1, the oxidation number ofeach hydrogen atom in the hydrogen molecule is 0. As specified by Rule6, the oxidation state of the hydrogen atom in the HCl molecule is +1; the

+1 −12HCl

0Cl2

0H2

+1 −1AgCl

+5 −2NO3

−+1Na+

+5 −2NO3

−+1

Ag+−1Cl−

+1Na+

+4 −2 +1 −2 +1 +4 −2SO2 + H2O ⎯→ H2SO3

C H A P T E R 1 9634

Photochromic LensesPhotochromic eyeglasses darken whenexposed to ultraviolet light and becometransparent again in the absence ofultraviolet light. This process is the resultof oxidation-reduction reactions. Silverchloride and copper(I) chloride areembedded in the lenses. The chlorideions absorb photons, and the silver chlo-ride dissociates and forms chlorineatoms and silver atoms. The elementalsilver darkens the lenses. Note that thechlorine ions are oxidized and the silveratoms are reduced. Then, the copper(I)ions reduce the chlorine atoms andform copper(II) ions. In the reverseprocess, the copper(II) ions oxidize the silver atoms back to the transpar-ent silver ions.

hydrogen atom is oxidized. Neither atom has totally lost or totally gainedany electrons. Hydrogen has donated a share of its bonding electron tothe chlorine but has not completely transferred that electron. The assign-ment of oxidation numbers allows an approximation of the electron dis-tribution of a molecule.An element can have different oxidation numbersin different compounds. This difference in oxidation numbers can revealthe difference in electron distribution of the compounds.

Reactants and products in redox reactions are not limited to monatom-ic ions and uncombined elements. Elements in molecular compounds orpolyatomic ions can also be oxidized and reduced if they have more thanone nonzero oxidation state.An example of this is provided in the reactionbetween the copper penny and nitric acid in which the nitrate ion, NO3

−, isconverted to nitrogen dioxide, NO2. Nitrogen is reduced in this reaction.Usually, we refer to the oxidation or reduction of the entire molecule orion. Instead of saying that the nitrogen atom is reduced, we say the nitrateion is reduced to nitrogen dioxide.

+5 +4. . . + NO3

− ⎯→ NO2 + . . .


1. How are oxidation numbers assigned?

2. Label each of the following half-reactions as eitheran oxidation or a reduction half-reaction:

a. + 2e− ⎯→ 2

b. ⎯→ + e−

c. 2 ⎯→ + 2e−

d. + 2e− ⎯→ 2

e. + e− ⎯→

f. ⎯→ + 2e−

g. + 2e− ⎯→

h. + e− ⎯→

3. Which of the following equations represent redoxreactions?a. 2KNO3(s ) ⎯→ 2KNO2(s ) + O2(g )

b. H2(g ) + CuO(s ) ⎯→ Cu(s ) + H2O(l )

c. NaOH(aq ) + HCl(aq ) ⎯→ NaCl(aq ) + H2O(l )

d. H2(g ) + Cl2(g ) ⎯→ 2HCl(g )

e. SO3(g ) + H2O(l ) ⎯→ H2SO4(aq )

4. For each redox equation identified in the previousquestion, determine which element is oxidized andwhich is reduced.

Critical Thinking

5. ANALYZING INFORMATION Use the followingequations for the redox reaction between Al3+ andNa to answer the questions below.

3 ⎯→ 3 + 3e− (oxidation)

+ 3e− ⎯→ (reduction)

3 + ⎯→ 3 + + (redox reaction)

a. Explain how this reaction illustrates that chargeis conserved in a redox reaction.

b. Explain how this reaction illustrates that massis conserved in a redox reaction.

c. Explain why electrons do not appear as react-ants or products in the combined equation.

0Al

+1Na

+3Al3+

0Na

0Al

+3Al3+

+1Na+

0Na

+2Fe2+

+3Fe3+

0Cu

+2Cu2+

+2Fe2+

0Fe

0Na

+1Na+

−1Cl−

0Cl2

0Cl2

−1Cl−

+1Na+

0Na

−1Br−

0Br2

SECTION REVIEW

C H A P T E R 1 9636

Skunk-Spray RemedyHave you ever given your pet atomato juice bath to get rid of thesmell of skunk spray on its coat?Chemistry has a much better way ofconquering skunk spray.

Paul Krebaum, the inventor of anew “deskunking” formula, saysthat while working as a materialsengineer, he constantly had to dealwith the less-than-pleasant smell ofthe hydrogen sulfide gas that wasreleased from one of his experi-ments. Venting off the gas only par-tially solved the problem. A bettersolution would be to eliminate thesmell entirely.

Mr. Krebaum rifled through his oldchemistry books and found thathydrogen peroxide could oxidizethese sulfur-containing compoundsto much less smelly components. Heimmediately whipped up a hydrogenperoxide mixture, and it worked likea charm.

The equation below shows thathydrogen sulfide reacts with H2O2 toform sulfate compounds that do nothave a bad odor.

2NaOH + 4H2O2 + H2S ⎯→Na2SO4 + 6H2O

“The receptors that are in yournose are sensitive to sulfur in its lowoxidation state,” says Mr. Krebaum.“However, they are not sensitive tosulfur in its high oxidation state.”

Some time later, a friend ofMr. Krebaum’s complained to himthat a skunk had sprayed his pet.

Because the odor in a skunk’s sprayalso comes from compounds con-taining sulfur in a low oxidationstate, Mr. Krebaum thought his solution might also work on thisage-old problem. He mixed up amilder version to try out on the pet: 1 qt of a 3% hydrogen peroxidesolution, 1/4 cup of baking soda, and1 tsp of liquid soap. His friend tried itout, and the result was one wet andunhappy—but much less smelly—pet.

Mr. Krebaum says that the hydro-gen peroxide in the remedy actuallyoxidizes the compounds, while thebaking soda reduces the acidity ofthe mixture and the soap helps towash out the greasy skunk spray.This reaction can be seen in the fol-lowing equation. The symbol R repre-sents all the other elements in the

sulfur-containing compound that is in skunk spray.

RSH + 3H2O2 + NaHCO3 ⎯→RSO3Na + 4H2O + CO2

The pet should be thoroughlywashed with the mixture, and careshould be taken to avoid the eyes. Ifthe mixture is left on for a few min-utes—long enough for the reactionto occur—and then rinsed away withtap water, the smell will disappear.

The formula does not bleach orcause any other negative sideeffects. Mr. Krebaum does have onewarning: Mix the formula just beforeusing it, because the mixture breaksdown quickly. The reaction releasesoxygen, so the formula should notbe kept in a sealed container.Pressure will build up, and the lidcould eventually blow off. For thisreason, bottles of “Krebaum’sSkunkinator” will not be appearingon drugstore shelves any time soon.

1. How did Paul Krebaum’sresearch into the properties ofH2S result in a benefit to dogowners?

2. What are some possible packag-ing designs that Paul Krebaumcould have used if he had wanted to sell his formula?

Questions

Skunk spray gets its odor from chemi-cals called mercaptans.


SECTION 2

OBJECTIVES

Explain what must be con-served in redox equations.

Balance redox equations by using the half-reactionmethod.

Balancing RedoxEquations

E quations for simple redox reactions can be balanced by inspection,which you learned to do in Chapter 8. Most redox equations, however,require more systematic methods. The equation-balancing processrequires the use of oxidation numbers. In a balanced equation, bothcharge and mass are conserved. Although oxidation and reduction half-reactions occur together, their reaction equations are balanced sepa-rately and then combined to give the balanced redox-reaction equation.

Half-Reaction Method

The half-reaction method, or ion-electron method, for balancing redoxequations consists of seven steps. Oxidation numbers are assigned to allatoms and polyatomic ions to determine which species are part of theredox process. The oxidation and reduction equations are balanced sep-arately for mass and charge. They are then added together to produce acomplete balanced equation. These seven steps are applied to balancethe reaction of hydrogen sulfide and nitric acid. Sulfuric acid, nitrogendioxide, and water are the products of the reaction.

1. Write the formula equation if it is not given in the problem. Thenwrite the ionic equation.

Formula equation: H2S + HNO3 ⎯→ H2SO4 + NO2 + H2O

Ionic equation: H2S + H+ + NO3− ⎯→ 2H+ + SO4

2− + NO2 + H2O

2. Assign oxidation numbers. Delete substances containing only el-ements that do not change oxidation state.

+ + ⎯→ + + +

The sulfur changes oxidation state from −2 to +6. The nitrogenchanges oxidation state from +5 to +4. The other substances aredeleted.

+ ⎯→ +

The remaining species are used in step 3.

+4 −2NO2

+6 −2SO4

2−+5 −2NO3

−+1 −2H2S

+1 −2H2O

+4 −2NO2

+6 −2SO4

2−+1

2H++5 −2NO3

−+1H+

+1 −2H2S

CROSS-DISCIPLINARYGo to go.hrw.com for a full-lengtharticle on redox reactions in photo-synthesis.

Keyword: HC6OXRX

3. Write the half-reaction for oxidation. In this example, the sulfur isbeing oxidized.

⎯→

• Balance the atoms. To balance the oxygen in this half-reaction, H2Omust be added to the left side. This gives 10 extra hydrogen atoms onthat side of the equation. Therefore, 10 hydrogen ions are added tothe right side. In basic solution, OH− ions and water may be used tobalance atoms.

+ 4H2O ⎯→ + 10H+

• Balance the charge. Electrons are added to the side having thegreater positive net charge. The left side of the equation has no netcharge; the right side has a net charge of 8+. For the charges to bal-ance, each side must have the same net charge.Therefore, 8 electronsare added to the product side so that it has no charge and balanceswith the reactant side of the equation. Notice that the oxidation ofsulfur from a state of −2 to +6 indicates a loss of 8 electrons.

+ 4H2O ⎯→ + 10H+ + 8e−

The oxidation half-reaction is now balanced.

4. Write the half-reaction for reduction. In this example, nitrogen isbeing reduced from a +5 state to a +4 state.

⎯→

• Balance the atoms. H2O must be added to the product side of thereaction to balance the oxygen atoms. Therefore, two hydrogen ionsmust be added to the reactant side to balance the hydrogen atoms.

+ 2H+ ⎯→ + H2O

• Balance the charge. Electrons are added to the side having thegreater positive net charge. The left side of the equation has a netcharge of 1+. Therefore, 1 electron must be added to this side to bal-ance the charge.

+ 2H+ + e− ⎯→ + H2O

The reduction half-reaction is now balanced.

5. Conserve charge by adjusting the coefficients in front of the elec-trons so that the number lost in oxidation equals the number gainedin reduction. Write the ratio of the number of electrons lost to thenumber of electrons gained.

=8

1

e− lost in oxidation

e− gained in reduction

+4NO2

+5NO3

−

+4NO2

+5NO3

−

+4NO2

+5NO3

−

+6SO4

2−−2

H2S

+6SO4

2−−2

H2S

+6SO4

2−−2

H2S

C H A P T E R 1 9638

This ratio is already in its lowest terms. If it were not, it would needto be reduced. Multiply the oxidation half-reaction by 1 (it remainsunchanged) and the reduction half-reaction by 8. The number ofelectrons lost now equals the number of electrons gained.

1� + 4H2O ⎯→ + 10H+ + 8e−�8 � + 2H+ + e− ⎯→ + H2O�

6. Combine the half-reactions, and cancel out anything common toboth sides of the equation.

+ 4H2O ⎯→ + 10H+ + 8e−

8 O3− + 16H+ + 8e− ⎯→ 8 O2 + 8H2O

+ + 8e− + + 4H2O ⎯→

+ + + 10H+ + 8e−

Each side of the above equation has 10H+, 8e−, and 4H2O. These can-cel each other out and do not appear in the balanced equation.

+ + 6H+ ⎯→ + 4H2O +

7. Combine ions to form the compounds shown in the original formu-la equation. Check to ensure that all other ions balance. The NO3

−

ion appeared as nitric acid in the original equation. There are only 6 hydrogen ions to pair with the 8 nitrate ions. Therefore, 2 hydrogenions must be added to complete this formula. If 2 hydrogen ions areadded to the left side of the equation, 2 hydrogen ions must also beadded to the right side of the equation.

8HNO3 + H2S ⎯→ 8NO2 + 4H2O + SO42− + 2H+

The sulfate ion appeared as sulfuric acid in the original equation.Thehydrogen ions added to the right side are used to complete the form-ula for sulfuric acid.

8HNO3 + H2S ⎯→ 8NO2 + 4H2O + H2SO4

A final check must be made to ensure that all elements are correctlybalanced.

+6SO4

2−+4

8NO2

−2H2S

+58NO3

−

+6SO4

2−48H2O

+48NO2

−2H2S

616H+

+58NO3

−

+4N

+5N

+6SO4

2−−2

H2S

+4NO2

+5NO3

−

+6SO4

2−−2

H2S


FIGURE 4 As a KMnO4 solutionis titrated into an acidic solution ofFeSO4, deep purple MnO4

− ions arereduced to colorless Mn2+ ions.When all Fe2+ ions are oxidized,MnO4

− ions are no longer reduced to colorless Mn2+ ions. Thus, the firstfaint appearance of the MnO4

− colorindicates the end point of thetitration.

SAMPLE PROBLEM A

Write a balanced equation for the reaction shown in Figure 4. A deep purple solution of potassium per-manganate is titrated with a colorless solution of iron(II) sulfate and sulfuric acid. The products areiron(III) sulfate, manganese(II) sulfate, potassium sulfate, and water—all of which are colorless.


KMnO4

FeSO4

in H2SO4

C H A P T E R 1 9640

1. Write the formula equation if it is not given in the problem. Then write the ionic equation.

KMnO4 + FeSO4 + H2SO4 ⎯→ Fe2(SO4)3 + MnSO4 + K2SO4 + H2OK+ + MnO4

− + Fe2+ + SO42− + 2H+ + SO4

2− ⎯→2Fe3+ + 3SO4

2− + Mn2+ + SO42− + 2K+ + SO4

2− + H2O

2. Assign oxidation numbers to each element and ion. Delete substances containing an ele-ment that does not change oxidation state.

Only ions or molecules whose oxidation numbers change are retained.

3. Write the half-reaction for oxidation. The iron shows the increase in oxidation number.Therefore, it is oxidized.

• Balance the mass. The mass is already balanced.

• Balance the charge.

4. Write the half-reaction for reduction. Manganese shows a change in oxidation numberfrom +7 to +2. It is reduced.

• Balance the mass. Water and hydrogen ions must be added to balance the oxygen atomsin the permanganate ion.

• Balance the charge.

5. Adjust the coefficients to conserve charge.

=

5(Fe2+ ⎯→ Fe3+ + e−)1(MnO4

− + 8H+ + 5e− ⎯→ Mn2+ + 4H2O)

1

5

e− lost in oxidation

e− gained in reduction

+7 +2MnO4

− + 8H+ + 5e− ⎯→ Mn2+ + 4H2O

+7 +2MnO4

− + 8H+ ⎯→ Mn2+ + 4H2O

+7 +2MnO4

− ⎯→ Mn2+

+2 +3Fe2+ ⎯→ Fe3+ + e−

+2 +3Fe2+ ⎯→ Fe3+

+7 −2 +2 +3 +2MnO4

− + Fe2+ ⎯→ Fe3+ + Mn2+

+3 +6 −2 +2 +6 −2 +1 +6 −2 +1 −22Fe3+ + 3SO4

2− + Mn2+ + SO42− + 2K+ + SO4

2− + H2O

+1 +7 −2 +2 +6 −2 +1 +6 −2K+ + MnO4

− + Fe2+ + SO42− + 2H+ + SO4

2− ⎯→

SOLUTION


Go to go.hrw.com formore practice problemsthat ask you to balanceredox equations.

Keyword: HC6OXRX

6. Combine the half-reactions and cancel.

5Fe2+ ⎯→ 5Fe3+ + 5e−

MnO4− + 8H+ + 5e− ⎯→ Mn2+ + 4H2O

MnO4− + 5Fe2+ + 8H+ + 5e− ⎯→ Mn2+ + 5Fe3+ + 4H2O + 5e−

7. Combine ions to form compounds from the original equation. The iron(III) productappears in the original equation as Fe2(SO4)3. Every iron(III) sulfate molecule requirestwo iron ions. Therefore, the entire equation must be multiplied by 2 to provide an evennumber of iron ions.

2(5Fe2+ + MnO4− + 8H+ ⎯→ 5Fe3+ + Mn2+ + 4H2O)

10Fe2+ + 2MnO4− + 16H+ ⎯→ 10Fe3+ + 2Mn2+ + 8H2O

The iron(II), iron(III), manganese(II), and 2 hydrogen ions in the original equation arepaired with sulfate ions. Iron(II) sulfate requires 10 sulfate ions, and sulfuric acidrequires 8 sulfate ions. To balance the equation, 18 sulfate ions must be added to eachside. On the product side, 15 of these ions form iron(III) sulfate, and 2 of them form man-ganese(II) sulfate. That leaves 1 sulfate ion unaccounted for. The permanganate ionrequires the addition of 2 potassium ions to each side. These 2 potassium ions formpotassium sulfate on the product side of the reaction.

10FeSO4 + 2KMnO4 + 8H2SO4 ⎯→ 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

Final inspection shows that atoms and charges are balanced.

1. Copper reacts with hot, concentrated sulfuric acid to form cop-per(II) sulfate, sulfur dioxide, and water. Write and balance theequation for this reaction.

2. Write and balance the equation for the reaction between nitricacid and potassium iodide. The products are potassium nitrate,iodine, nitrogen monoxide, and water.


1. What two quantities are conserved in redoxequations?

2. Why do we add H+ and H2O to some half-reactions and OH− and H2O to others?

3. Balance the following redox reaction:

Na2SnO2 + Bi(OH)3 ⎯→ Bi + Na2SnO3 + H2O

Critical Thinking

4. RELATING IDEAS When heated, elemental phos-phorus, P4, produces phosphine, PH3, and phos-phoric acid, H3PO4. How many grams of phosphineare produced if 56 g P4 have reacted?

SECTION REVIEW

C H A P T E R 1 9642

SECTION 3 Oxidizing andReducing Agents

A reducing agent is a substance that has the potential to cause anothersubstance to be reduced. Reducing agents lose electrons; they attain amore positive oxidation state during an oxidation-reduction reaction.Therefore, the reducing agent is the oxidized substance.

An oxidizing agent is a substance that has the potential to cause an-other substance to be oxidized. Oxidizing agents gain electrons andattain a more negative oxidation state during an oxidation-reductionreaction. The oxidizing agent is the reduced substance. Table 2 helpsclarify the terms describing the oxidation-reduction process.

Strengths of Oxidizing and Reducing Agents

Different substances can be compared and rated by their relative potential as reducing and oxidizing agents. For example, the order of theelements in the activity series (see Table 3 on page 286) is related to eachelement’s tendency to lose electrons. Elements in this series lose elec-trons to the positively charged ions of any element below them in theseries. The more active an element is, the greater its tendency to loseelectrons and the better a reducing agent it is. The greater the distanceis between two elements in the list, the more likely it is that a redoxreaction will take place between them.

These elements and some other familiar substances are arranged inTable 3 according to their activity as oxidizing and reducing agents. Thefluorine atom is the most highly electronegative atom. It is also the most

OBJECTIVES

Relate chemical activity tooxidizing and reducingstrength.

Explain the concept of disproportionation.

Change in Change inTerm oxidation number electron population

Oxidation in a positive direction loss of electrons

Reduction in a negative direction gain of electrons

Oxidizing agent in a negative direction gains electrons

Reducing agent in a positive direction loses electrons

TABLE 2 Oxidation-Reduction Terminology

active oxidizing agent. Because of its strong attraction for its own elec-trons, the fluoride ion is the weakest reducing agent.The negative ion ofa strong oxidizing agent is a weak reducing agent.

The positive ion of a strong reducing agent is a weak oxidizing agent.As shown in Table 3, Li atoms are strong reducing agents because Li isa very active metal.When Li atoms oxidize, they produce Li+ ions, whichare unlikely to reacquire electrons, so Li+ ions are weak oxidizingagents.

The left column of each pair also shows the relative abilities of met-als listed in the table to displace other metals from their compounds.Zinc, for example, appears above copper. Thus, zinc is the more activereducing agent, and it displaces copper ions from solutions of coppercompounds, as illustrated in Figure 5. The copper(II) ion, on the otherhand, is a more active oxidizing agent than the zinc ion.

Nonmetals and some important ions also are included in the series inTable 3. Any reducing agent is oxidized by the oxidizing agents belowit. Observe that F2 displaces Cl−, Br−, and I− ions from their solutions.Cl2 displaces Br− and I− ions, and Br2 displaces I− ions.The equation forthe displacement of Br− by Cl2 is as follows.

Cl2 + 2Br−(aq) ⎯→ 2Cl−(aq) + Br2

2 ⎯→ + 2e− (oxidation)

+ 2e− ⎯→ 2 (reduction)

In every redox reaction, there is one reducing agent and one oxi-dizing agent. In the preceding example, Br− is the reducing agent andCl2 is the oxidizing agent.

−1Cl−

0Cl2

0Br2

−1Br−


FIGURE 5 Zinc displaces copper ions from a copper(II) sulfate solution.Metallic copper precipitates.

Reducing Oxidizingagents agents

Li Li+

K K+

Ca Ca2+

Na Na+

Mg Mg2+

Al Al3+

Zn Zn2+

Cr Cr3+

Fe Fe2+

Ni Ni2+

Sn Sn2+

Pb Pb2+

H2 H3O+

H2S S

Cu Cu2+

I− I2

MnO42− MnO4

−

Fe2+ Fe3+

Hg Hg22+

Ag Ag+

NO2− NO3

−

Br− Br2

Mn2+ MnO2

SO2 H2SO4 (conc.)

Cr3+ Cr2O72−

Cl− Cl2Mn2+ MnO4

−

F − F2

TABLE 3 Relative Strengthof Oxidizing and ReducingAgents

Incr

easi

ng s

tren

gth

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯→

Increasing strength⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

⎯⎯

→

Disproportionation

Some substances can be both reduced and oxidized easily. For example,peroxide ions, O2

2−, have a relatively unstable covalent bond betweenthe two oxygen atoms. The electron-dot formula is written as follows.

Each oxygen atom has an oxidation number of −1. The peroxide ionstructure represents an intermediate oxidation state between O2 andO2−. Therefore, the peroxide ion is highly reactive.

Hydrogen peroxide, H2O2, is a covalent compound. It decomposesinto water and molecular oxygen, as shown in the equation below.

2H2 ⎯→ 2H2 +0O2

−2O

−1O2

O2−

O

C H A P T E R 1 9644

Materials• aluminum foil• beaker, 250 mL• 1 M copper(II) chloride solu-

tion, CuCl2• 3% hydrogen peroxide• manganese dioxide• metric ruler• scissors• test-tube clamp• test tube, 16 × 150 mm• wooden splint

ProcedureRecord all of your results in a data table.

1. Put 10 mL of hydrogen perox-ide in a test tube, and add asmall amount of manganesedioxide (equal to the size ofabout half a pea). What is theresult?

2. Insert a glowing wooden splintinto the test tube (see dia-gram). What is the result? Ifoxygen is produced, a glowingwooden splint inserted into thetest tube will glow brighter.

3. Fill the 250 mL beaker halfwaywith the copper(II) chloridesolution.

4. Cut foil into 2 cm × 12 cmstrips.

5. Add the aluminum strips to thecopper(II) chloride solution.Use a glass rod to stir the mix-ture, and observe for 12 to15 minutes. What is the result?

Discussion

1. Write balanced equationsshowing what happened ineach of the reactions.

2. Write a conclusion for the twoexperiments.

Redox Reactions

Wear safety goggles and an apron.

H2O2

Woodensplint

Clamp

Notice that in this reaction, hydrogen peroxide is both oxidized andreduced. Oxygen atoms that become part of gaseous oxygen moleculesare oxidized. The oxidation number of these oxygen atoms increasesfrom −1 to 0. Oxygen atoms that become part of water are reduced. Theoxidation number of these oxygen atoms decreases from −1 to −2. Aprocess in which a substance acts as both an oxidizing agent and a reduc-ing agent is called disproportionation. A substance that undergoes dis-proportionation is both self-oxidizing and self-reducing.

The bombardier beetle defends itself by spraying its enemies with anunpleasant hot chemical mixture, as shown in Figure 6. The catalyzeddisproportionation of hydrogen peroxide produces hot oxygen gas. Thisgas gives the insect an ability to eject irritating chemicals from itsabdomen with explosive force.


1. Describe the chemical activity of the alkali metalsand of the halogens on the basis of oxidizing andreducing strength.

2. The photo on the left depicts two redox reactions.Both nails are in a sulfuric acid solution. Answerthe following questions:a. When zinc is wrapped around an iron nail,

is the iron or zinc oxidized?b. When copper is wrapped around an iron

nail, is the iron or copper oxidized?

3. Would Cl2 be reduced by I−? Explain.

4. Which is the stronger oxidizing agent in each ofthe following pairs: Cu2+ or Al3+, I2 or S, F2 or Li+?

5. What is meant by disproportionation?

Critical Thinking

6. ORGANIZING IDEAS In general, where in the periodic table are the elements found that in elemental form are the strongest oxidizingagents? Explain.

SECTION REVIEW

FIGURE 6 A bombardier beetlecan repel large predators such asfrogs with a chemical defense mech-anism that uses the disproportiona-tion of hydrogen peroxide.


C H A P T E R 1 9646

• Charge and mass are conserved in a balanced redox equation.• In the half-reaction method for balancing equations, the atoms

and charge of oxidation and reduction equations are balancedseparately. Then, they are combined to give a complete balancedequation.

• In a half-reaction, the charge on the reactant side must equalthe charge on the product side, but these charges do not needto be zero.

• For the half-reaction method, the atoms in each half-reactionare balanced by adding H+ ions and H2O molecules in acidicsolutions. If the solution is basic, OH− ions and H2O moleculesare added to balance the atoms in each half-reaction.

• The number of electrons lost in the oxidation half-reactionmust equal the number of electrons gained in the reductionhalf-reaction. The two half-reactions must be multiplied byappropriate factors to ensure that the same number of elec-trons are transferred.

• Oxidation numbers are assigned by the set of rules listed inTable 1. Oxidation numbers are based on the distribution ofelectrons in a molecule.

• Oxidation-reduction reactions consist of two half-reactionsthat must occur simultaneously.

• Oxidation-reduction reactions are identified by examining thechanges in the oxidation numbers of atoms in the reactantsand products.

• Oxidation involves the loss of electrons, and reduction involvesthe gain of electrons.

• A species whose oxidation number increases is oxidized. Aspecies whose oxidation number decreases is reduced.

oxidationoxidizedreductionreducedoxidation-reduction reactionredox reactionhalf-reaction

Vocabulary

Oxidation and Reduction

Balancing Redox Equations

• The substance that is reduced in redox reactions is the oxidiz-ing agent because it acquires electrons from the substance thatis oxidized.

• The substance that is oxidized in a redox reaction is the reducingagent because it supplies the electrons to the substance that isreduced.

• Strong reducing agents are substances that easily give up elec-trons.

• Disproportionation is a process in which a substance is both anoxidizing agent and a reducing agent.

reducing agentoxidizing agentdisproportionation

Vocabulary

Oxidizing and Reducing Agents

C H A P T E R R E V I E WFor more practice, go to the Problem Bank in Appendix D.


Oxidation and ReductionSECTION 1 REVIEW

1. a. Distinguish between the processes of oxida-tion and reduction.

b. Write an equation to illustrate each process.2. Which of the following are redox reactions?

a. 2Na + Cl2 ⎯→ 2NaClb. C + O2 ⎯→ CO2c. 2H2O ⎯→ 2H2 + O2d. NaCl + AgNO3 ⎯→ AgCl + NaNO3e. NH3 + HCl ⎯→ NH4

+ + Cl−

f. 2KClO3 ⎯→ 2KCl + 3O2g. H2 + Cl2 ⎯→ 2HClh. H2SO4 + 2KOH ⎯→ K2SO4 + 2H2Oi. Zn + CuSO4 ⎯→ ZnSO4 + Cu

3. For each oxidation-reduction reaction in theprevious question, identify what is oxidized andwhat is reduced.

PRACTICE PROBLEMS

4. Each of the following atom/ion pairs undergoesthe oxidation number change indicated below.For each pair, determine whether oxidation orreduction has occurred, and then write the elec-tronic equation indicating the correspondingnumber of electrons lost or gained.a. K ⎯→ K+ e. H2 ⎯→ H+

b. S ⎯→ S2− f. O2 ⎯→ O2−

c. Mg ⎯→ Mg2+ g. Fe3+ ⎯→ Fe2+

d. F− ⎯→ F2 h. Mn2+ ⎯→ MnO4–

5. Identify the following reactions as redox ornonredox:a. 2NH4Cl(aq) + Ca(OH)2(aq) ⎯→

2NH3(aq) + 2H2O(l) + CaCl2(aq)b. 2HNO3(aq) + 3H2S(g) ⎯→

2NO(g) + 4H2O(l) + 3S(s)c. [Be(H2O)4]2+(aq) + H2O(l) ⎯→

H3O+(aq) + [Be(H2O)3OH]+(aq)6. Arrange the following in order of increasing

oxidation number of the xenon atom: CsXeF8,Xe, XeF2, XeOF2, XeO3, and XeF.

7. Determine the oxidation number of each atomindicated in the following:a. H2 f. HNO3b. H2O g. H2SO4c. Al h. Ca(OH)2d. MgO i. Fe(NO3)2e. Al2S3 j. O2

Balancing Redox EquationsSECTION 2 REVIEW

8. Label the following half-reactions as eitherreduction or oxidation half-reactions.a. H2S ⎯→ S + 2e− + 2H+

b. SO2 + 4e− + 2H2O ⎯→ S + 4OH−

c. ClO3− + 6H+ + 6e− ⎯→ Cl− + 3H2O

d. Mn(CN)64− ⎯→ Mn(CN)6

3− + e−

9. What are the oxidation states of the elementsthat changed oxidation states in the half-reactions in the above question?

10. Balance the equation for the following reactionin a basic solution. Give balanced equations forboth half-reactions and the balanced equationfor the overall reaction.KMnO4 + NaIO3 ⎯→ MnO2 + NaIO4

PRACTICE PROBLEMS

11. For each requested step, use the half-reactionmethod to balance the oxidation-reductionequation below. (Hint: See Sample Problem A.)

K + H2O ⎯→ KOH + H2a. Write the ionic equation, and assign oxida-

tion numbers to all atoms to determine whatis oxidized and what is reduced.

b. Write the equation for the reduction, and balance it for both atoms and charge.

c. Write the equation for the oxidation, and balance it for both atoms and charge.

d. Multiply the coefficients of the oxidation andreduction equations so that the number ofelectrons lost equals the number of electronsgained. Add the two equations.

e. Add species as necessary to balance the overall formula equation.

CHAPTER REVIEW

12. Use the method in the previous problem to bal-ance each of the reactions below.a. HI + HNO2 ⎯→ NO + I2 + H2Ob. FeCl3 + H2S ⎯→ FeCl2 + HCl + S

13. Balance the equation for the reaction in which hot, concentrated sulfuric acid reacts withzinc to form zinc sulfate, hydrogen sulfide,and water.

Oxidizing and Reducing AgentsSECTION 3 REVIEW

14. a. Identify the most active reducing agentamong all common elements.

b. Why are all of the elements in its group in theperiodic table very active reducing agents?

c. Identify the most active oxidizing agentamong the common elements.

15. Use Table 3 to identify the strongest and weak-est reducing agents among the substances listedwithin each of the following groupings:a. Ca, Ag, Sn, Cl−

b. Fe, Hg, Al, Br−

c. F−, Pb, Mn2+, Na16. Use Table 3 to respond to each of the following:

a. Would Al be oxidized by Ni2+?b. Would Cu be oxidized by Ag+?c. Would Pb be oxidized by Na+?d. Would F2 be reduced by Cl−?e. Would Br2 be reduced by Cl−?

17. Identify the following reactions as redox ornonredox:a. Mg(s) + ZnCl2(aq) ⎯→ Zn(s) + MgCl2(aq)b. 2H2(g) + OF2(g) ⎯→ H2O(g) + 2HF(g)c. 2KI(aq) + Pb(NO3)2(aq) ⎯→

PbI2(s) + 2KNO3(aq)d. CaO(s) + H2O(l) ⎯→ Ca(OH)2(aq)e. 3CuCl2(aq) + 2(NH4)3PO4(aq) ⎯→

6NH4Cl(aq) + Cu3(PO4)2(s)f. CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(g)

MIXED REVIEW

18. Arrange the following in order of decreasingoxidation number of the nitrogen atom: N2,NH3, N2O4, N2O, N2H4, and NO3

−.19. Balance the following redox equations:

a. SbCl5 + KI ⎯→ KCl + I2 + SbCl3b. Ca(OH)2 + NaOH + ClO2 + C ⎯→

NaClO2 + CaCO3 + H2O20. Balance the following equations in basic

solution:a. PbO2 + KCl ⎯→ KClO + KPb(OH)3b. KMnO4 + KIO3 ⎯→ MnO2 + KIO4c. K2MnO4 ⎯→ MnO2 + KMnO4

21. Balance the following equations in acidic solution:a. MnO4

− + Cl− ⎯→ Mn2+ + HClOb. NO3

− + I2 ⎯→ IO3− + NO2

c. NO2− ⎯→ NO + NO3

−

22. Interpreting Graphics Given the activity tablebelow, determine whether a reaction will occur ornot. If the reaction will occur, give the products.

a. L and M+

b. P and M+

c. P and T+

23. Drawing Conclusions A substance has an ele-ment in one of its highest possible oxidationstates. Is this substance more likely to be an oxi-dizing agent or a reducing agent? Explain yourreasoning.

ReducingAgents

OxidizingAgents

LMPT

L+

M+

P+

T+

CRITICAL THINKING

C H A P T E R 1 9648

CHAPTER REVIEW


24. Drawing Conclusions Use Table 3 to decide ifa redox reaction would occur between the twospecies, and if so, write the balanced equation.Explain your reasoning.a. Cl2 and Br2b. Sn2+ and Zn

25. Drawing Conclusions An element that dispro-portionates must have at least how many differ-ent oxidation states? Explain your reasoning.

26. Several reactions of aluminum are shown in thecommon reactions section for Group 13 of theElements Handbook. Use these reactions toanswer the following:a. Which of the five reactions shown are

oxidation-reduction reactions? How do you know?

b. For each redox reaction you listed in item a,identify what is oxidized and what is reduced.

c. Write half-reactions for each equation youlisted in item a.

27. Aluminum is described in Group 13 of theElements Handbook as a self-protecting metal.This property of aluminum results from a redoxreaction.a. Write the redox equation for the oxidation of

aluminum.b. Write the half-reactions for this reaction, and

show the number of electrons transferred.c. What problems are associated with the

buildup of aluminum oxide on electricalwiring made of aluminum?

USING THE HANDBOOK

28. Oxidizing agents are used in the cleaning indus-try. Research three different oxidizing agentsused in this area, and write a report on the advan-tages and disadvantages of these compounds.

29. Oxidizing and reducing agents play importantroles in biological systems. Research the role ofone of these agents in a biological process.Write a report describing the process and therole of oxidation and reduction.

30. Boilers are used to convert water to steam inelectric power plants. Dissolved oxygen in thewater promotes corrosion of the steel used inboiler parts. Explain how dissolved oxygen isremoved from the water in boilers.

31. Performance For one day, record situationsthat show evidence of oxidation-reduction reac-tions. Identify the reactants and the products,and determine whether there is proof that achemical reaction has taken place.


RESEARCH & WRITING

Problem Solving S-

To balance redox equations for reactions in basic solution:

• Add OH− and H2O to balance oxygen and hydrogen in the redox half-reactions.

• Add OH− ions to the side of the equation that needs oxygen atoms. Make sureyou add enough OH− ions so that the number of oxygen atoms added is twice thenumber needed.

• Then, add enough H2O molecules to the other side of the equation to balance thehydrogen atoms.

SAMPLE

A redox equation must conserve both mass and charge. So, to balance a redox equation,you must balance both atoms and charge (electrons). The problem-solving tips and sam-ple below show how to balance an equation for a redox reaction in basic solution.

The following unbalanced equation represents a redox reaction that takes place in a basicsolution containing KOH. Balance the redox equation.

Br2(l) + KOH(aq) ⎯→ KBr(aq) + KBrO3(aq)Write the full ionic equation, assign oxidation numbers, and eliminate species whose oxi-

dation numbers do not change. The result is the following equation:

B0

r2 ⎯→ Br−−1

+ B+5

rO3−

Divide this equation into half-reactions. Note that Br2 is the reactant in both half-reactions.Reduction: Br2 ⎯→ Br− Oxidation: Br2 ⎯→ BrO3

−

Add H2O and OH− to balance atoms in basic solution.Then, add electrons to balance charge.Reduction: Br2 + 2e− ⎯→ 2Br− (no need to add H2O or OH−)Oxidation: 12OH− + Br2 ⎯→ 2BrO3

− + 6H2O + 10e−

To balance transferred electrons, you must multiply the reduction half-reaction by 5 so thatboth reactions have 10e−.

5 × (Br2 + 2e− ⎯→ 2Br−) = 5Br2 + 10e− ⎯→ 10Br−

Combining the two half-reactions gives5Br2 + 12OH− + Br2 + 10e− ⎯→ 10Br− + 2BrO3

− + 6H2O + 10e−

Canceling common species gives6Br2 + 12OH− ⎯→ 10Br− + 2BrO3

− + 6H2OReturning the potassium ions to the equation gives

6Br2 + 12KOH ⎯→ 10KBr + 2KBrO3 + 6H2O, or 3Br2 + 6KOH ⎯→5KBr + KBrO3 + 3H2O

C H A P T E R 1 9650

Math Tutor BALANCING REDOX EQUATIONS

1. Balance the following equation for a redoxreaction that takes place in basic solution:MnO2(s) + NaClO3(aq) + NaOH(aq) ⎯→

NaMnO4(aq) + NaCl(aq) + H2O(l)

2. Balance the following equation for a redoxreaction that takes place in basic solution:N2O(g) + KClO(aq) + KOH(aq) ⎯→

KCl(aq) + KNO2(aq) + H2O(l)

PRACTICE PROBLEMS



MULTIPLE CHOICE

1.In the following reaction, which species isreduced?

2K + Br2 ⎯→ 2K+ + 2Br−

A. K B. Br2C. All of the aboveD. None of the above

2.The oxidation number of the sulfur atom in theSO4

2− ion isA. +2.B. −2.C. +6.D. +4.

3.A half-reactionA. involves a change in the oxidation state of an

element.B. always contains H2O molecules.C. always contains H+ ions.D. All of the above

4.In the following reaction, which is the oxidizingagent?AgNO2 + Cl2 + 2KOH ⎯→

AgNO3 + 2KCl + H2OA. AgNO2B. Cl2C. KOHD. KCl

5.What are the oxidation states (in increasingorder) of the element that undergoes dispropor-tionation in the following reaction:

Cl2 + H2O ⎯→ HCl + HOClA. −1, 0, +2B. −1, 0, +1C. −2, −1, 0D. None of the above

6.Which reaction is a redox reaction?A. Al2O3 + 6HCl ⎯→ 2AlCl3 + 3H2OB. 2HCO3

− ⎯→ CO2 + CO32− + H2O

C. SiBr4 + 3H2O ⎯→ H2SiO3 + 4HBrD. H2O + PbO2 + NaOH + KCl ⎯→

KClO + NaPb(OH)3

7.Arrange the following in order of increasingoxidation number of the sulfur atom: S2O3

2−,S4O6

2−, HSO4−, and H2S.

A. H2S, S2O32−, S4O6

2−, HSO4−

B. S2O32−, H2S, S4O6

2−, HSO4−

C. H2S, S2O32−, HSO4

−, S4O62−

D. HSO4−, S2O3

2−, S4O62−, H2S

8.Which answer contains the correct informationabout the following reaction:

2Pb(NO3)2 ⎯→ 2PbO + 4NO2 + O2A. This reaction is a decomposition reaction and

not a redox reaction.B. This reaction is a redox reaction in which the

lead is reduced and the oxygen is oxidized.C. This reaction is a disproportionation reaction.D. This reaction is a redox reaction in which the

nitrogen is reduced and the oxygen is oxidized.

SHORT ANSWER

9.Determine the oxidation numbers for Cu in thesuperconductor YBa2Cu3O7. Yttrium (Y) has anoxidation number of +3. (Cu does not have oxi-dation numbers greater than +3.) Give onlyinteger oxidation numbers.

10.What is an oxidizing agent?

EXTENDED RESPONSE

11.B, F, K, and L are four unknown reducingagents that oxidize to singly charged cations.Using the following information, construct atable showing the relative strengths of the oxi-dizing and reducing agents. Data: F reduces K+,B+, and L+. B+ oxidizes K and F, but not L.

12.Balance the equation for the following reactionin basic solution:

ClO2 ⎯→ KClO3 + KClO2

Give the balanced equation for each half-reaction and for the overall reaction. Give theoxidizing agent and the reducing agent.


If you are short on time, quicklyscan the unanswered questions to see which might be easiest to answer.

BACKGROUNDIn Chapter 15, you studied acid-base titrations inwhich an unknown amount of acid is titrated with acarefully measured amount of base. In this procedure,a similar approach called a redox titration is used. In aredox titration, the reducing agent, Fe2+, is oxidized toFe3+ by the oxidizing agent, MnO4

−. When this processoccurs, the Mn in MnO4

− changes from a +7 to a +2oxidation state and has a noticeably different color.You can use this color change to signify a redox reac-tion “end point.” When the reaction is complete, anyexcess MnO4

− added to the reaction mixture will givethe solution a pink or purple color. The volume datafrom the titration, the known molarity of the KMnO4solution, and the mole ratio from the balanced redoxequation will give you the information you need tocalculate the molarity of the FeSO4 solution.

SAFETY

For review of safety, please see Safety in theChemistry Laboratory in the front of your book.

PREPARATION1. In your lab notebook, prepare a data table like

the one shown on the next page.

2. Clean two 50 mL burets with a buret brush anddistilled water. Rinse each buret at least threetimes with distilled water to remove contaminants.

3. Label one 250 mL beaker “0.0200 M KMnO4”and the other “FeSO4.” Label three of the flasks“1,” “2,” and “3.” Label the 400 mL beaker“Waste.” Label one buret “KMnO4” and the other“FeSO4.”

OBJECTIVES

• Demonstrate proficiency in performingredox titrations and recognizing end pointsof a redox reaction.

• Write a balanced oxidation-reduction equation for a redox reaction.

• Determine the concentration of a solutionby using stoichiometry and volume datafrom a titration.

MATERIALS

• 0.0200 M KMnO4

• 1.0 M H2SO4

• 100 mL graduated cylinder

• 125 mL Erlenmeyer flasks, 4

• 250 mL beakers, 2

• 400 mL beaker

• burets, 2

• distilled water

• double buret clamp

• FeSO4 solution

• ring stand

• wash bottle

Reduction of Manganese in Permanganate Ion

CHAPTER LABSEE PRE-LAB

VOLUMETRIC ANALYSISMICROL A B

C H A P T E R 1 9652

2. Slowly add KMnO4 from the buret to the FeSO4in the flask while swirling the flask. When thecolor of the solution matches the color standardyou prepared in Preparation step 8, record inyour data table the final readings of the burets.

3. Empty the titration flask into the waste beaker.Repeat the titration procedure in steps 1 and 2with the flasks labeled “2” and “3.”

CLEANUP AND DISPOSAL4. Dispose of the contents of the waste

beaker in the container designated byyour teacher. Also, pour the color-standard flask into this container. Wash yourhands thoroughly after cleaning up the area and equipment.

ANALYSIS AND INTERPRETATION1. Organizing Ideas: Write the balanced equation

for the redox reaction of FeSO4 and KMnO4.

2. Evaluating Data: Calculate the number of molesof MnO4

− reduced in each trial.

3. Analyzing Information: Calculate the number of moles of Fe2+ oxidized in each trial.

4. Applying Conclusions: Calculate the averageconcentration (molarity) of the iron(II) sulfate solution.

EXTENSIONS1. Designing Experiments: What possible sources

of error can you identify with this procedure? If you can think of ways to eliminate them, askyour teacher to approve your plan, and run theprocedure again.

4. Measure approximately 75 mL of 0.0200 MKMnO4, and pour it into the appropriatelylabeled beaker. Obtain approximately 75 mL of FeSO4 solution, and pour it into the appro-priately labeled beaker.

5. Rinse one buret three times with a few millilitersof 0.0200 M KMnO4 from the appropriatelylabeled beaker. Collect these rinses in the wastebeaker. Rinse the other buret three times withsmall amounts of FeSO4 solution from theappropriately labeled beaker. Collect these rinses in the waste beaker.

6. Set up the burets as instructed by your teacher.Fill one buret with approximately 50 mL of0.0200 M KMnO4 from the beaker, and fill theother buret with approximately 50 mL of theFeSO4 solution from the other beaker.

7. With the waste beaker underneath its tip, openthe KMnO4 buret long enough to be sure theburet tip is filled. Repeat the process for theFeSO4 buret.

8. Add 50 mL of distilled water to one of the125 mL Erlenmeyer flasks, and add one drop of 0.0200 M KMnO4 to the flask. Set this mix-ture aside to use as a color standard. It can becompared with the titration mixture to deter-mine the end point.

PROCEDURE1. Record in your data table the initial buret read-

ings for both solutions. Add 10 mL of thehydrated iron(II) sulfate solution, FeSO4•7H2O,to the flask labeled “1.” Add 5 mL of 1 M H2SO4 to the FeSO4 solution in this flask.The acid will help keep the Fe2+ ions in thereduced state, which will allow you time totitrate.


Trial Initial KMnO4 Final KMnO4 Initial FeSO4 Final FeSO4volume (mL) volume (mL) volume (mL) volume (mL)

1

23

DATA TABLE

chapter 17 reaction kinetics...reaction kinetics 561 section 1 o bjectives explain the concept of...

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